Sum of the first natural numbers: how many and what are the most common methods to verify it? [duplicate]Proof for formula for sum of sequence $1+2+3+ldots+n$?Spivak's Calculus (Chapter 2, Exercise 17)What are the most famous (common used) precalculus books and its differences?Compute $ax^3+by^3$ given $ax^2+by^2$ and $ax+by$.Calculate the Gauss integral without squaring it firstSum of first n natural numbers proofPerforming Induction on the process of Induction (Function Spaces?)Summation of series of first $2n$ natural numbers and first $3n$ natural numbersProof by induction (Recursion)sum of the first $n^2$ natural numbers closed form

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Sum of the first natural numbers: how many and what are the most common methods to verify it? [duplicate]


Proof for formula for sum of sequence $1+2+3+ldots+n$?Spivak's Calculus (Chapter 2, Exercise 17)What are the most famous (common used) precalculus books and its differences?Compute $ax^3+by^3$ given $ax^2+by^2$ and $ax+by$.Calculate the Gauss integral without squaring it firstSum of first n natural numbers proofPerforming Induction on the process of Induction (Function Spaces?)Summation of series of first $2n$ natural numbers and first $3n$ natural numbersProof by induction (Recursion)sum of the first $n^2$ natural numbers closed form






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;

.everyonelovesstackoverflowposition:absolute;height:1px;width:1px;opacity:0;top:0;left:0;pointer-events:none;








3















$begingroup$



This question already has an answer here:



  • Proof for formula for sum of sequence $1+2+3+ldots+n$?

    32 answers



We know that Gauss has shown that the sum $S$ of the first $n$ natural numbers is given by the relation:



$$S=fracn(n+1)2 tag*$$
The proof that I remember most frequently is as follows:



Let be $S=1+2+dotsb+(n-1)+n tag1$ We can write $S$ it also as: $tag2 S=n+(n-1)+dotsb+2+1.$
By adding up member to member we get:
$tag3 2S=underbrace(n+1)+(n+1)+dotsb+2+1_n-mathrmtimes.$
Hence we obtain the $(^ast)$.



How many other simple methods exist to calculate the sum of the first natural numbers?










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$endgroup$





marked as duplicate by Aloizio Macedo Sep 27 at 6:42


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    Induction. $$
    $endgroup$
    – copper.hat
    Sep 26 at 21:24










  • $begingroup$
    @copper.hat Yes, of course. But I remember that there is another simple form which has the similar form of my question.
    $endgroup$
    – Sebastiano
    Sep 26 at 21:26

















3















$begingroup$



This question already has an answer here:



  • Proof for formula for sum of sequence $1+2+3+ldots+n$?

    32 answers



We know that Gauss has shown that the sum $S$ of the first $n$ natural numbers is given by the relation:



$$S=fracn(n+1)2 tag*$$
The proof that I remember most frequently is as follows:



Let be $S=1+2+dotsb+(n-1)+n tag1$ We can write $S$ it also as: $tag2 S=n+(n-1)+dotsb+2+1.$
By adding up member to member we get:
$tag3 2S=underbrace(n+1)+(n+1)+dotsb+2+1_n-mathrmtimes.$
Hence we obtain the $(^ast)$.



How many other simple methods exist to calculate the sum of the first natural numbers?










share|cite|improve this question









$endgroup$





marked as duplicate by Aloizio Macedo Sep 27 at 6:42


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    Induction. $$
    $endgroup$
    – copper.hat
    Sep 26 at 21:24










  • $begingroup$
    @copper.hat Yes, of course. But I remember that there is another simple form which has the similar form of my question.
    $endgroup$
    – Sebastiano
    Sep 26 at 21:26













3













3









3


2



$begingroup$



This question already has an answer here:



  • Proof for formula for sum of sequence $1+2+3+ldots+n$?

    32 answers



We know that Gauss has shown that the sum $S$ of the first $n$ natural numbers is given by the relation:



$$S=fracn(n+1)2 tag*$$
The proof that I remember most frequently is as follows:



Let be $S=1+2+dotsb+(n-1)+n tag1$ We can write $S$ it also as: $tag2 S=n+(n-1)+dotsb+2+1.$
By adding up member to member we get:
$tag3 2S=underbrace(n+1)+(n+1)+dotsb+2+1_n-mathrmtimes.$
Hence we obtain the $(^ast)$.



How many other simple methods exist to calculate the sum of the first natural numbers?










share|cite|improve this question









$endgroup$





This question already has an answer here:



  • Proof for formula for sum of sequence $1+2+3+ldots+n$?

    32 answers



We know that Gauss has shown that the sum $S$ of the first $n$ natural numbers is given by the relation:



$$S=fracn(n+1)2 tag*$$
The proof that I remember most frequently is as follows:



Let be $S=1+2+dotsb+(n-1)+n tag1$ We can write $S$ it also as: $tag2 S=n+(n-1)+dotsb+2+1.$
By adding up member to member we get:
$tag3 2S=underbrace(n+1)+(n+1)+dotsb+2+1_n-mathrmtimes.$
Hence we obtain the $(^ast)$.



How many other simple methods exist to calculate the sum of the first natural numbers?





This question already has an answer here:



  • Proof for formula for sum of sequence $1+2+3+ldots+n$?

    32 answers







algebra-precalculus induction alternative-proof






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 26 at 21:20









SebastianoSebastiano

6192 silver badges18 bronze badges




6192 silver badges18 bronze badges





marked as duplicate by Aloizio Macedo Sep 27 at 6:42


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











marked as duplicate by Aloizio Macedo Sep 27 at 6:42


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Aloizio Macedo Sep 27 at 6:42


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • $begingroup$
    Induction. $$
    $endgroup$
    – copper.hat
    Sep 26 at 21:24










  • $begingroup$
    @copper.hat Yes, of course. But I remember that there is another simple form which has the similar form of my question.
    $endgroup$
    – Sebastiano
    Sep 26 at 21:26
















  • $begingroup$
    Induction. $$
    $endgroup$
    – copper.hat
    Sep 26 at 21:24










  • $begingroup$
    @copper.hat Yes, of course. But I remember that there is another simple form which has the similar form of my question.
    $endgroup$
    – Sebastiano
    Sep 26 at 21:26















$begingroup$
Induction. $$
$endgroup$
– copper.hat
Sep 26 at 21:24




$begingroup$
Induction. $$
$endgroup$
– copper.hat
Sep 26 at 21:24












$begingroup$
@copper.hat Yes, of course. But I remember that there is another simple form which has the similar form of my question.
$endgroup$
– Sebastiano
Sep 26 at 21:26




$begingroup$
@copper.hat Yes, of course. But I remember that there is another simple form which has the similar form of my question.
$endgroup$
– Sebastiano
Sep 26 at 21:26










5 Answers
5






active

oldest

votes


















4

















$begingroup$

Here is a proof using ideas from probability theory. Let $X$ be a random variable that is uniformly distributed on the set $1,dotsc, n$. Then it is easy to see that $n+1-X$ has the same distribution as $X$ whence
$$
E(n+1-X)=EX
$$

i.e. $n+1-EX=EX$ so
$$
EX=fracn+12tag0.
$$

But on the other hand,
$$
EX=sum_k=1^n kP(X=k)=frac1nsum_k=1^n k.tag1
$$

Combining $(0)$ and $(1)$ we get that
$$
sum_k=1^nk=fracn(n+1)2.
$$






share|cite|improve this answer










$endgroup$













  • $begingroup$
    Thank you for all users, and also for you. My sincere "grazie".
    $endgroup$
    – Sebastiano
    Sep 26 at 22:17


















2

















$begingroup$

Certainly overkill, but we can do this with the method of generating functions: let
$$f(x)=sum_k=0^nx^n=frac1-x^n+11-x$$
So that $f'(1)=sum_k=0^nk$. We just need to find $f'(x)$, and we proceed as usual:
$$fracddxfrac1-x^n+11-x=frac-(n+1)x^n(1-x)+(1-x^n+1)(1-x)^2$$
Taking the limit as $xto1$, we have by L'hopital's rule and after simplification,
$$lim_xto1frac-n(n+1)x^n-1(1-x)-2(1-x)=lim_xto1fracn(n+1)2=fracn(n+1)2$$
From which we obtain the conclusion.






share|cite|improve this answer










$endgroup$













  • $begingroup$
    If you’re going with generating functions, why not instead compute $[x^n-1](1-x)^-3$? That only requires using the Generalized Binomial Theorem.
    $endgroup$
    – amd
    Sep 28 at 7:29










  • $begingroup$
    @amd I never said you couldn't do that, of course. That would be another way.
    $endgroup$
    – YiFan
    Sep 28 at 7:33


















2

















$begingroup$

The expression of $S_n$ must be a quadratic polynomial (because the difference $S_n-S_n-1=n$ is a linear polynomial). As $S_0=0$, it is of the form $an^2+bn.$



By identification,



$$begincasesS_1=a+b=1,\S_2=4a+2b=3endcases$$



and $$a=b=dfrac12.$$




The Lagrangian polynomial by $(0,0)$, $(1,1)$, $(2,3)$ is $dfracx^22+dfrac x2$.




$$S(n)-S(n-1)=an^2+bn-a(n-1)^2-b(n-1)=a(2n-1)+b=n$$



so that



$$a=b=dfrac12.$$




A posteriori:



$$S_n-S_n-1=fracn(n+1)2-frac(n-1)n2=n.$$




As an arithmetic progression is linear, the average term is the average of the extreme terms.



$$fracS_nn=frac1+n2$$ and $$S_n=fracn(n+1)2.$$




Let the function defined by a geometric series



$$f(x):=sum_k=0^n x^k=fracx^n+1-1x-1.$$



We differentiate on $x$,



$$f'(x)=sum_k=0^n k x^k-1=fracx^n+1-1x-1=frac(n+1)x^n(x-1)-(x^n+1-1)(x-1)^2.$$



Then with $x=1$, using L'Hospital,



$$f'(1)=sum_k=0^n k =lim_xto0fracnx^n+1-(n+1)x^n+1(x-1)^2 =lim_xto0fracn(n+1)x^n-(n+1)nx^n-12(x-1)=fracn(n+1)2.$$






share|cite|improve this answer












$endgroup$













  • $begingroup$
    Thank you very much also to you. Have you used the telescopic rule? :o)
    $endgroup$
    – Sebastiano
    Sep 26 at 21:37






  • 1




    $begingroup$
    @Sebastiano The telescoping can be discussed in terms of binomial coefficients, viz.$$sum_k=1^nk=sum_k=1^nbinomk1=binomn+12$$by the hockey-stick identity. But we can also derive that without telescoping by a double-counting argument.
    $endgroup$
    – J.G.
    Sep 26 at 21:42











  • $begingroup$
    @J.G. Very difficult for my students to understand. However thank you for your precious comment.
    $endgroup$
    – Sebastiano
    Sep 26 at 21:45






  • 1




    $begingroup$
    @Sebastiano The double-counting argument I mentioned might be the easiest way for them to understand it.
    $endgroup$
    – J.G.
    Sep 26 at 21:48


















1

















$begingroup$

The identity



$$sum_k=1^nk=fracn(n+1)2$$



is a classical result which can be easily proved by the following



enter image description here






share|cite|improve this answer










$endgroup$





















    1

















    $begingroup$

    Use can use the sum of AP to arrive at the same conclusion.
    $a=1, d=1$
    $$S_n = nover 2(2a+(n-1)d)$$
    $$S_n = nover 2(2+(n-1))$$
    $$S_n = nover 2(n+1)$$






    share|cite|improve this answer










    $endgroup$













    • $begingroup$
      Sam. Excuse me very much. What is the word AP? :-(
      $endgroup$
      – Sebastiano
      Sep 26 at 21:40






    • 1




      $begingroup$
      @Sebastiano Arithmetic progression.
      $endgroup$
      – J.G.
      Sep 26 at 21:41


















    5 Answers
    5






    active

    oldest

    votes








    5 Answers
    5






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4

















    $begingroup$

    Here is a proof using ideas from probability theory. Let $X$ be a random variable that is uniformly distributed on the set $1,dotsc, n$. Then it is easy to see that $n+1-X$ has the same distribution as $X$ whence
    $$
    E(n+1-X)=EX
    $$

    i.e. $n+1-EX=EX$ so
    $$
    EX=fracn+12tag0.
    $$

    But on the other hand,
    $$
    EX=sum_k=1^n kP(X=k)=frac1nsum_k=1^n k.tag1
    $$

    Combining $(0)$ and $(1)$ we get that
    $$
    sum_k=1^nk=fracn(n+1)2.
    $$






    share|cite|improve this answer










    $endgroup$













    • $begingroup$
      Thank you for all users, and also for you. My sincere "grazie".
      $endgroup$
      – Sebastiano
      Sep 26 at 22:17















    4

















    $begingroup$

    Here is a proof using ideas from probability theory. Let $X$ be a random variable that is uniformly distributed on the set $1,dotsc, n$. Then it is easy to see that $n+1-X$ has the same distribution as $X$ whence
    $$
    E(n+1-X)=EX
    $$

    i.e. $n+1-EX=EX$ so
    $$
    EX=fracn+12tag0.
    $$

    But on the other hand,
    $$
    EX=sum_k=1^n kP(X=k)=frac1nsum_k=1^n k.tag1
    $$

    Combining $(0)$ and $(1)$ we get that
    $$
    sum_k=1^nk=fracn(n+1)2.
    $$






    share|cite|improve this answer










    $endgroup$













    • $begingroup$
      Thank you for all users, and also for you. My sincere "grazie".
      $endgroup$
      – Sebastiano
      Sep 26 at 22:17













    4















    4











    4







    $begingroup$

    Here is a proof using ideas from probability theory. Let $X$ be a random variable that is uniformly distributed on the set $1,dotsc, n$. Then it is easy to see that $n+1-X$ has the same distribution as $X$ whence
    $$
    E(n+1-X)=EX
    $$

    i.e. $n+1-EX=EX$ so
    $$
    EX=fracn+12tag0.
    $$

    But on the other hand,
    $$
    EX=sum_k=1^n kP(X=k)=frac1nsum_k=1^n k.tag1
    $$

    Combining $(0)$ and $(1)$ we get that
    $$
    sum_k=1^nk=fracn(n+1)2.
    $$






    share|cite|improve this answer










    $endgroup$



    Here is a proof using ideas from probability theory. Let $X$ be a random variable that is uniformly distributed on the set $1,dotsc, n$. Then it is easy to see that $n+1-X$ has the same distribution as $X$ whence
    $$
    E(n+1-X)=EX
    $$

    i.e. $n+1-EX=EX$ so
    $$
    EX=fracn+12tag0.
    $$

    But on the other hand,
    $$
    EX=sum_k=1^n kP(X=k)=frac1nsum_k=1^n k.tag1
    $$

    Combining $(0)$ and $(1)$ we get that
    $$
    sum_k=1^nk=fracn(n+1)2.
    $$







    share|cite|improve this answer













    share|cite|improve this answer




    share|cite|improve this answer










    answered Sep 26 at 22:15









    Foobaz JohnFoobaz John

    26.4k4 gold badges15 silver badges57 bronze badges




    26.4k4 gold badges15 silver badges57 bronze badges














    • $begingroup$
      Thank you for all users, and also for you. My sincere "grazie".
      $endgroup$
      – Sebastiano
      Sep 26 at 22:17
















    • $begingroup$
      Thank you for all users, and also for you. My sincere "grazie".
      $endgroup$
      – Sebastiano
      Sep 26 at 22:17















    $begingroup$
    Thank you for all users, and also for you. My sincere "grazie".
    $endgroup$
    – Sebastiano
    Sep 26 at 22:17




    $begingroup$
    Thank you for all users, and also for you. My sincere "grazie".
    $endgroup$
    – Sebastiano
    Sep 26 at 22:17













    2

















    $begingroup$

    Certainly overkill, but we can do this with the method of generating functions: let
    $$f(x)=sum_k=0^nx^n=frac1-x^n+11-x$$
    So that $f'(1)=sum_k=0^nk$. We just need to find $f'(x)$, and we proceed as usual:
    $$fracddxfrac1-x^n+11-x=frac-(n+1)x^n(1-x)+(1-x^n+1)(1-x)^2$$
    Taking the limit as $xto1$, we have by L'hopital's rule and after simplification,
    $$lim_xto1frac-n(n+1)x^n-1(1-x)-2(1-x)=lim_xto1fracn(n+1)2=fracn(n+1)2$$
    From which we obtain the conclusion.






    share|cite|improve this answer










    $endgroup$













    • $begingroup$
      If you’re going with generating functions, why not instead compute $[x^n-1](1-x)^-3$? That only requires using the Generalized Binomial Theorem.
      $endgroup$
      – amd
      Sep 28 at 7:29










    • $begingroup$
      @amd I never said you couldn't do that, of course. That would be another way.
      $endgroup$
      – YiFan
      Sep 28 at 7:33















    2

















    $begingroup$

    Certainly overkill, but we can do this with the method of generating functions: let
    $$f(x)=sum_k=0^nx^n=frac1-x^n+11-x$$
    So that $f'(1)=sum_k=0^nk$. We just need to find $f'(x)$, and we proceed as usual:
    $$fracddxfrac1-x^n+11-x=frac-(n+1)x^n(1-x)+(1-x^n+1)(1-x)^2$$
    Taking the limit as $xto1$, we have by L'hopital's rule and after simplification,
    $$lim_xto1frac-n(n+1)x^n-1(1-x)-2(1-x)=lim_xto1fracn(n+1)2=fracn(n+1)2$$
    From which we obtain the conclusion.






    share|cite|improve this answer










    $endgroup$













    • $begingroup$
      If you’re going with generating functions, why not instead compute $[x^n-1](1-x)^-3$? That only requires using the Generalized Binomial Theorem.
      $endgroup$
      – amd
      Sep 28 at 7:29










    • $begingroup$
      @amd I never said you couldn't do that, of course. That would be another way.
      $endgroup$
      – YiFan
      Sep 28 at 7:33













    2















    2











    2







    $begingroup$

    Certainly overkill, but we can do this with the method of generating functions: let
    $$f(x)=sum_k=0^nx^n=frac1-x^n+11-x$$
    So that $f'(1)=sum_k=0^nk$. We just need to find $f'(x)$, and we proceed as usual:
    $$fracddxfrac1-x^n+11-x=frac-(n+1)x^n(1-x)+(1-x^n+1)(1-x)^2$$
    Taking the limit as $xto1$, we have by L'hopital's rule and after simplification,
    $$lim_xto1frac-n(n+1)x^n-1(1-x)-2(1-x)=lim_xto1fracn(n+1)2=fracn(n+1)2$$
    From which we obtain the conclusion.






    share|cite|improve this answer










    $endgroup$



    Certainly overkill, but we can do this with the method of generating functions: let
    $$f(x)=sum_k=0^nx^n=frac1-x^n+11-x$$
    So that $f'(1)=sum_k=0^nk$. We just need to find $f'(x)$, and we proceed as usual:
    $$fracddxfrac1-x^n+11-x=frac-(n+1)x^n(1-x)+(1-x^n+1)(1-x)^2$$
    Taking the limit as $xto1$, we have by L'hopital's rule and after simplification,
    $$lim_xto1frac-n(n+1)x^n-1(1-x)-2(1-x)=lim_xto1fracn(n+1)2=fracn(n+1)2$$
    From which we obtain the conclusion.







    share|cite|improve this answer













    share|cite|improve this answer




    share|cite|improve this answer










    answered Sep 27 at 6:03









    YiFanYiFan

    11.2k4 gold badges16 silver badges43 bronze badges




    11.2k4 gold badges16 silver badges43 bronze badges














    • $begingroup$
      If you’re going with generating functions, why not instead compute $[x^n-1](1-x)^-3$? That only requires using the Generalized Binomial Theorem.
      $endgroup$
      – amd
      Sep 28 at 7:29










    • $begingroup$
      @amd I never said you couldn't do that, of course. That would be another way.
      $endgroup$
      – YiFan
      Sep 28 at 7:33
















    • $begingroup$
      If you’re going with generating functions, why not instead compute $[x^n-1](1-x)^-3$? That only requires using the Generalized Binomial Theorem.
      $endgroup$
      – amd
      Sep 28 at 7:29










    • $begingroup$
      @amd I never said you couldn't do that, of course. That would be another way.
      $endgroup$
      – YiFan
      Sep 28 at 7:33















    $begingroup$
    If you’re going with generating functions, why not instead compute $[x^n-1](1-x)^-3$? That only requires using the Generalized Binomial Theorem.
    $endgroup$
    – amd
    Sep 28 at 7:29




    $begingroup$
    If you’re going with generating functions, why not instead compute $[x^n-1](1-x)^-3$? That only requires using the Generalized Binomial Theorem.
    $endgroup$
    – amd
    Sep 28 at 7:29












    $begingroup$
    @amd I never said you couldn't do that, of course. That would be another way.
    $endgroup$
    – YiFan
    Sep 28 at 7:33




    $begingroup$
    @amd I never said you couldn't do that, of course. That would be another way.
    $endgroup$
    – YiFan
    Sep 28 at 7:33











    2

















    $begingroup$

    The expression of $S_n$ must be a quadratic polynomial (because the difference $S_n-S_n-1=n$ is a linear polynomial). As $S_0=0$, it is of the form $an^2+bn.$



    By identification,



    $$begincasesS_1=a+b=1,\S_2=4a+2b=3endcases$$



    and $$a=b=dfrac12.$$




    The Lagrangian polynomial by $(0,0)$, $(1,1)$, $(2,3)$ is $dfracx^22+dfrac x2$.




    $$S(n)-S(n-1)=an^2+bn-a(n-1)^2-b(n-1)=a(2n-1)+b=n$$



    so that



    $$a=b=dfrac12.$$




    A posteriori:



    $$S_n-S_n-1=fracn(n+1)2-frac(n-1)n2=n.$$




    As an arithmetic progression is linear, the average term is the average of the extreme terms.



    $$fracS_nn=frac1+n2$$ and $$S_n=fracn(n+1)2.$$




    Let the function defined by a geometric series



    $$f(x):=sum_k=0^n x^k=fracx^n+1-1x-1.$$



    We differentiate on $x$,



    $$f'(x)=sum_k=0^n k x^k-1=fracx^n+1-1x-1=frac(n+1)x^n(x-1)-(x^n+1-1)(x-1)^2.$$



    Then with $x=1$, using L'Hospital,



    $$f'(1)=sum_k=0^n k =lim_xto0fracnx^n+1-(n+1)x^n+1(x-1)^2 =lim_xto0fracn(n+1)x^n-(n+1)nx^n-12(x-1)=fracn(n+1)2.$$






    share|cite|improve this answer












    $endgroup$













    • $begingroup$
      Thank you very much also to you. Have you used the telescopic rule? :o)
      $endgroup$
      – Sebastiano
      Sep 26 at 21:37






    • 1




      $begingroup$
      @Sebastiano The telescoping can be discussed in terms of binomial coefficients, viz.$$sum_k=1^nk=sum_k=1^nbinomk1=binomn+12$$by the hockey-stick identity. But we can also derive that without telescoping by a double-counting argument.
      $endgroup$
      – J.G.
      Sep 26 at 21:42











    • $begingroup$
      @J.G. Very difficult for my students to understand. However thank you for your precious comment.
      $endgroup$
      – Sebastiano
      Sep 26 at 21:45






    • 1




      $begingroup$
      @Sebastiano The double-counting argument I mentioned might be the easiest way for them to understand it.
      $endgroup$
      – J.G.
      Sep 26 at 21:48















    2

















    $begingroup$

    The expression of $S_n$ must be a quadratic polynomial (because the difference $S_n-S_n-1=n$ is a linear polynomial). As $S_0=0$, it is of the form $an^2+bn.$



    By identification,



    $$begincasesS_1=a+b=1,\S_2=4a+2b=3endcases$$



    and $$a=b=dfrac12.$$




    The Lagrangian polynomial by $(0,0)$, $(1,1)$, $(2,3)$ is $dfracx^22+dfrac x2$.




    $$S(n)-S(n-1)=an^2+bn-a(n-1)^2-b(n-1)=a(2n-1)+b=n$$



    so that



    $$a=b=dfrac12.$$




    A posteriori:



    $$S_n-S_n-1=fracn(n+1)2-frac(n-1)n2=n.$$




    As an arithmetic progression is linear, the average term is the average of the extreme terms.



    $$fracS_nn=frac1+n2$$ and $$S_n=fracn(n+1)2.$$




    Let the function defined by a geometric series



    $$f(x):=sum_k=0^n x^k=fracx^n+1-1x-1.$$



    We differentiate on $x$,



    $$f'(x)=sum_k=0^n k x^k-1=fracx^n+1-1x-1=frac(n+1)x^n(x-1)-(x^n+1-1)(x-1)^2.$$



    Then with $x=1$, using L'Hospital,



    $$f'(1)=sum_k=0^n k =lim_xto0fracnx^n+1-(n+1)x^n+1(x-1)^2 =lim_xto0fracn(n+1)x^n-(n+1)nx^n-12(x-1)=fracn(n+1)2.$$






    share|cite|improve this answer












    $endgroup$













    • $begingroup$
      Thank you very much also to you. Have you used the telescopic rule? :o)
      $endgroup$
      – Sebastiano
      Sep 26 at 21:37






    • 1




      $begingroup$
      @Sebastiano The telescoping can be discussed in terms of binomial coefficients, viz.$$sum_k=1^nk=sum_k=1^nbinomk1=binomn+12$$by the hockey-stick identity. But we can also derive that without telescoping by a double-counting argument.
      $endgroup$
      – J.G.
      Sep 26 at 21:42











    • $begingroup$
      @J.G. Very difficult for my students to understand. However thank you for your precious comment.
      $endgroup$
      – Sebastiano
      Sep 26 at 21:45






    • 1




      $begingroup$
      @Sebastiano The double-counting argument I mentioned might be the easiest way for them to understand it.
      $endgroup$
      – J.G.
      Sep 26 at 21:48













    2















    2











    2







    $begingroup$

    The expression of $S_n$ must be a quadratic polynomial (because the difference $S_n-S_n-1=n$ is a linear polynomial). As $S_0=0$, it is of the form $an^2+bn.$



    By identification,



    $$begincasesS_1=a+b=1,\S_2=4a+2b=3endcases$$



    and $$a=b=dfrac12.$$




    The Lagrangian polynomial by $(0,0)$, $(1,1)$, $(2,3)$ is $dfracx^22+dfrac x2$.




    $$S(n)-S(n-1)=an^2+bn-a(n-1)^2-b(n-1)=a(2n-1)+b=n$$



    so that



    $$a=b=dfrac12.$$




    A posteriori:



    $$S_n-S_n-1=fracn(n+1)2-frac(n-1)n2=n.$$




    As an arithmetic progression is linear, the average term is the average of the extreme terms.



    $$fracS_nn=frac1+n2$$ and $$S_n=fracn(n+1)2.$$




    Let the function defined by a geometric series



    $$f(x):=sum_k=0^n x^k=fracx^n+1-1x-1.$$



    We differentiate on $x$,



    $$f'(x)=sum_k=0^n k x^k-1=fracx^n+1-1x-1=frac(n+1)x^n(x-1)-(x^n+1-1)(x-1)^2.$$



    Then with $x=1$, using L'Hospital,



    $$f'(1)=sum_k=0^n k =lim_xto0fracnx^n+1-(n+1)x^n+1(x-1)^2 =lim_xto0fracn(n+1)x^n-(n+1)nx^n-12(x-1)=fracn(n+1)2.$$






    share|cite|improve this answer












    $endgroup$



    The expression of $S_n$ must be a quadratic polynomial (because the difference $S_n-S_n-1=n$ is a linear polynomial). As $S_0=0$, it is of the form $an^2+bn.$



    By identification,



    $$begincasesS_1=a+b=1,\S_2=4a+2b=3endcases$$



    and $$a=b=dfrac12.$$




    The Lagrangian polynomial by $(0,0)$, $(1,1)$, $(2,3)$ is $dfracx^22+dfrac x2$.




    $$S(n)-S(n-1)=an^2+bn-a(n-1)^2-b(n-1)=a(2n-1)+b=n$$



    so that



    $$a=b=dfrac12.$$




    A posteriori:



    $$S_n-S_n-1=fracn(n+1)2-frac(n-1)n2=n.$$




    As an arithmetic progression is linear, the average term is the average of the extreme terms.



    $$fracS_nn=frac1+n2$$ and $$S_n=fracn(n+1)2.$$




    Let the function defined by a geometric series



    $$f(x):=sum_k=0^n x^k=fracx^n+1-1x-1.$$



    We differentiate on $x$,



    $$f'(x)=sum_k=0^n k x^k-1=fracx^n+1-1x-1=frac(n+1)x^n(x-1)-(x^n+1-1)(x-1)^2.$$



    Then with $x=1$, using L'Hospital,



    $$f'(1)=sum_k=0^n k =lim_xto0fracnx^n+1-(n+1)x^n+1(x-1)^2 =lim_xto0fracn(n+1)x^n-(n+1)nx^n-12(x-1)=fracn(n+1)2.$$







    share|cite|improve this answer















    share|cite|improve this answer




    share|cite|improve this answer








    edited Sep 27 at 8:33

























    answered Sep 26 at 21:36









    Yves DaoustYves Daoust

    152k13 gold badges92 silver badges251 bronze badges




    152k13 gold badges92 silver badges251 bronze badges














    • $begingroup$
      Thank you very much also to you. Have you used the telescopic rule? :o)
      $endgroup$
      – Sebastiano
      Sep 26 at 21:37






    • 1




      $begingroup$
      @Sebastiano The telescoping can be discussed in terms of binomial coefficients, viz.$$sum_k=1^nk=sum_k=1^nbinomk1=binomn+12$$by the hockey-stick identity. But we can also derive that without telescoping by a double-counting argument.
      $endgroup$
      – J.G.
      Sep 26 at 21:42











    • $begingroup$
      @J.G. Very difficult for my students to understand. However thank you for your precious comment.
      $endgroup$
      – Sebastiano
      Sep 26 at 21:45






    • 1




      $begingroup$
      @Sebastiano The double-counting argument I mentioned might be the easiest way for them to understand it.
      $endgroup$
      – J.G.
      Sep 26 at 21:48
















    • $begingroup$
      Thank you very much also to you. Have you used the telescopic rule? :o)
      $endgroup$
      – Sebastiano
      Sep 26 at 21:37






    • 1




      $begingroup$
      @Sebastiano The telescoping can be discussed in terms of binomial coefficients, viz.$$sum_k=1^nk=sum_k=1^nbinomk1=binomn+12$$by the hockey-stick identity. But we can also derive that without telescoping by a double-counting argument.
      $endgroup$
      – J.G.
      Sep 26 at 21:42











    • $begingroup$
      @J.G. Very difficult for my students to understand. However thank you for your precious comment.
      $endgroup$
      – Sebastiano
      Sep 26 at 21:45






    • 1




      $begingroup$
      @Sebastiano The double-counting argument I mentioned might be the easiest way for them to understand it.
      $endgroup$
      – J.G.
      Sep 26 at 21:48















    $begingroup$
    Thank you very much also to you. Have you used the telescopic rule? :o)
    $endgroup$
    – Sebastiano
    Sep 26 at 21:37




    $begingroup$
    Thank you very much also to you. Have you used the telescopic rule? :o)
    $endgroup$
    – Sebastiano
    Sep 26 at 21:37




    1




    1




    $begingroup$
    @Sebastiano The telescoping can be discussed in terms of binomial coefficients, viz.$$sum_k=1^nk=sum_k=1^nbinomk1=binomn+12$$by the hockey-stick identity. But we can also derive that without telescoping by a double-counting argument.
    $endgroup$
    – J.G.
    Sep 26 at 21:42





    $begingroup$
    @Sebastiano The telescoping can be discussed in terms of binomial coefficients, viz.$$sum_k=1^nk=sum_k=1^nbinomk1=binomn+12$$by the hockey-stick identity. But we can also derive that without telescoping by a double-counting argument.
    $endgroup$
    – J.G.
    Sep 26 at 21:42













    $begingroup$
    @J.G. Very difficult for my students to understand. However thank you for your precious comment.
    $endgroup$
    – Sebastiano
    Sep 26 at 21:45




    $begingroup$
    @J.G. Very difficult for my students to understand. However thank you for your precious comment.
    $endgroup$
    – Sebastiano
    Sep 26 at 21:45




    1




    1




    $begingroup$
    @Sebastiano The double-counting argument I mentioned might be the easiest way for them to understand it.
    $endgroup$
    – J.G.
    Sep 26 at 21:48




    $begingroup$
    @Sebastiano The double-counting argument I mentioned might be the easiest way for them to understand it.
    $endgroup$
    – J.G.
    Sep 26 at 21:48











    1

















    $begingroup$

    The identity



    $$sum_k=1^nk=fracn(n+1)2$$



    is a classical result which can be easily proved by the following



    enter image description here






    share|cite|improve this answer










    $endgroup$


















      1

















      $begingroup$

      The identity



      $$sum_k=1^nk=fracn(n+1)2$$



      is a classical result which can be easily proved by the following



      enter image description here






      share|cite|improve this answer










      $endgroup$
















        1















        1











        1







        $begingroup$

        The identity



        $$sum_k=1^nk=fracn(n+1)2$$



        is a classical result which can be easily proved by the following



        enter image description here






        share|cite|improve this answer










        $endgroup$



        The identity



        $$sum_k=1^nk=fracn(n+1)2$$



        is a classical result which can be easily proved by the following



        enter image description here







        share|cite|improve this answer













        share|cite|improve this answer




        share|cite|improve this answer










        answered Sep 26 at 21:31









        useruser

        111k10 gold badges49 silver badges102 bronze badges




        111k10 gold badges49 silver badges102 bronze badges
























            1

















            $begingroup$

            Use can use the sum of AP to arrive at the same conclusion.
            $a=1, d=1$
            $$S_n = nover 2(2a+(n-1)d)$$
            $$S_n = nover 2(2+(n-1))$$
            $$S_n = nover 2(n+1)$$






            share|cite|improve this answer










            $endgroup$













            • $begingroup$
              Sam. Excuse me very much. What is the word AP? :-(
              $endgroup$
              – Sebastiano
              Sep 26 at 21:40






            • 1




              $begingroup$
              @Sebastiano Arithmetic progression.
              $endgroup$
              – J.G.
              Sep 26 at 21:41















            1

















            $begingroup$

            Use can use the sum of AP to arrive at the same conclusion.
            $a=1, d=1$
            $$S_n = nover 2(2a+(n-1)d)$$
            $$S_n = nover 2(2+(n-1))$$
            $$S_n = nover 2(n+1)$$






            share|cite|improve this answer










            $endgroup$













            • $begingroup$
              Sam. Excuse me very much. What is the word AP? :-(
              $endgroup$
              – Sebastiano
              Sep 26 at 21:40






            • 1




              $begingroup$
              @Sebastiano Arithmetic progression.
              $endgroup$
              – J.G.
              Sep 26 at 21:41













            1















            1











            1







            $begingroup$

            Use can use the sum of AP to arrive at the same conclusion.
            $a=1, d=1$
            $$S_n = nover 2(2a+(n-1)d)$$
            $$S_n = nover 2(2+(n-1))$$
            $$S_n = nover 2(n+1)$$






            share|cite|improve this answer










            $endgroup$



            Use can use the sum of AP to arrive at the same conclusion.
            $a=1, d=1$
            $$S_n = nover 2(2a+(n-1)d)$$
            $$S_n = nover 2(2+(n-1))$$
            $$S_n = nover 2(n+1)$$







            share|cite|improve this answer













            share|cite|improve this answer




            share|cite|improve this answer










            answered Sep 26 at 21:37









            SamSam

            9111 silver badge11 bronze badges




            9111 silver badge11 bronze badges














            • $begingroup$
              Sam. Excuse me very much. What is the word AP? :-(
              $endgroup$
              – Sebastiano
              Sep 26 at 21:40






            • 1




              $begingroup$
              @Sebastiano Arithmetic progression.
              $endgroup$
              – J.G.
              Sep 26 at 21:41
















            • $begingroup$
              Sam. Excuse me very much. What is the word AP? :-(
              $endgroup$
              – Sebastiano
              Sep 26 at 21:40






            • 1




              $begingroup$
              @Sebastiano Arithmetic progression.
              $endgroup$
              – J.G.
              Sep 26 at 21:41















            $begingroup$
            Sam. Excuse me very much. What is the word AP? :-(
            $endgroup$
            – Sebastiano
            Sep 26 at 21:40




            $begingroup$
            Sam. Excuse me very much. What is the word AP? :-(
            $endgroup$
            – Sebastiano
            Sep 26 at 21:40




            1




            1




            $begingroup$
            @Sebastiano Arithmetic progression.
            $endgroup$
            – J.G.
            Sep 26 at 21:41




            $begingroup$
            @Sebastiano Arithmetic progression.
            $endgroup$
            – J.G.
            Sep 26 at 21:41



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