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I was teaching $$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\\\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}$$ for the first I did direct telescopic proof like below $$\sum_{k=1}^{n}k=\sum_{k=1}^{n}k(\frac{k+1}{2}-\frac{k-1}{2})=\sum_{k=1}^{n}(\frac{k(k+1)}{2}-\frac{k(k-1}{2}))=\\\sum_{k=1}^{n}(f(k)-f(k-1))=\frac{n(n+1)}{2}-0$$ for the second I did a classic proof $$\begin{align} \sum_{k=1}^n k^2 & = \frac{1}{3}(n+1)^3 - \frac{1}{2}n(n+1) - \frac{1}{3}(n+1) \\ & = \frac{1}{6}(n+1) \left[ 2(n+1)^2 - 3n - 2\right] \\ & = \frac{1}{6}(n+1)(2n^2 +n) \\ & = \frac{1}{6}n(n+1)(2n+1) \end{align}$$ some of the student asked for direct telescopic proof for the case... I can't find this kind of proof. can anybody help me to find, or write this kind proving. \

I tried to rewrite $1=\frac{k+1}{2}-\frac{k-1}{2}$and I have $$\sum_{k=1}^{n}k^2(\frac{k+1}{2}-\frac{k-1}{2})=$$ I can't go further more.

I promised to my students to try to find a direct proof Idea. Thanks for any help.

Khosrotash
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3 Answers3

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Use binomial coefficients. Write $k^2=2\binom{k}{2}+\binom{k}{1}$ so $\sum_{k=1}^nk^2=\sum_{k=1}^n(a_k-a_{k-1})$ with $$a_n=2\binom{n+1}{3}+\binom{n+1}{2}=\frac13 n(n^2-1)+\frac12 n(n+1)=\frac16 n(n+1)(2n+1).$$

J.G.
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    Not to hijack your post but I think it is worth mentioning that (1) Every polynomial $p(k)$ in $k$ can be written as a sum of binomial coefficient polynomials $\binom{k}{n}$. (2) The sum $\sum_{k=1}^n \binom{k}{a}$ is given by $\binom{n+1}{a+1}$, this is the so-called hockey stick identity (and can be thought of as a telescoping sum). Combining these two facts allows you to quickly compute (and prove) closed formulas for all sums of the form $\sum_{k=1}^n p(k)$. – Nate Sep 25 '19 at 19:44
  • @Nate Indeed. – J.G. Sep 25 '19 at 19:45
  • Using the hockey stick identity is, indeed, the standard way of producing these sums. It is very clean. My alternative approach of integrating the previous power sum function produces a bit nasty looking correction terms. Integration has the nice side feature that we avoid the need to expand the monomial $n^\ell$ as a linear combination of the binomial coefficients $\binom n t, 0<t\le \ell$. – Jyrki Lahtonen Sep 27 '19 at 10:06
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Noting that $$ (n+1)^2-n^2=2n+1, (n+1)^3-n^3=3n^2+3n+1 $$ one has $$ n=\frac12\bigg[(n+1)^2-n^2\bigg]-\frac12,n^2=\frac13\bigg[(n+1)^3-n^3\bigg]-n-\frac13. $$ So $$ n^2=\frac13\bigg[(n+1)^3-n^3\bigg]-n-\frac13=\frac13\bigg[(n+1)^3-n^3\bigg]-\frac12\bigg[(n+1)^2-n^2\bigg]+\frac16 $$ and hence \begin{eqnarray} \sum_{k=1}^nk^2&=&\sum_{k=1}^n\left[\frac13\bigg[(k+1)^3-k^3\bigg]-\frac12\bigg[(k+1)^2-k^2\bigg]+\frac16\right]\\ &=&\frac13\bigg[(n+1)^3-1\bigg]-\frac12\bigg[(n+1)^2-1\bigg]+\frac n6\\ &=&\frac16n(n+1)(2n+1) \end{eqnarray}

xpaul
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I quite endorse the other answers. Proffering the following alternative approach to producing correct "telescoping" functions for I find it kinda neat.

Your telescoping argument used the fact that for $f(x)=x(x+1)/2$ we have the identity $$ f(x)-f(x-1)=x.\tag{1} $$ The goal is to find another function $F(x)$ satisfying a similar identity $F(x)-F(x-1)=x^2$ instead of $(1)$.

If we integrate both sides of the identity $(1)$ we get the identity $$ \begin{aligned} \frac12x^2&=\int_{t=0}^xt\,dt\\ &=\int_{t=0}^x(f(t)-f(t-1))\,dt\\ &=\int_{t=0}^xf(t)\,dt-\int_{t=-1}^{x-1}f(t)\,dt\\ &=\int_{t=0}^xf(t)\,dt-\int_{t=0}^{x-1}f(t)\,dt-\int_{t=-1}^0f(t)\,dt\\ &=g(x)-g(x-1)+\frac1{12}, \end{aligned} \tag{2} $$ where $$ g(x)=\int_{t=0}^xf(t)\,dt=\frac16x^3+\frac14x^2. $$ Two problems here. We got $\dfrac12x^2$ instead of $x^2$ as the difference. But it is easy to fix that by multiplying everything with two. The other problem is that "integration constant $C=1/12$".

To deal with that pesky $1/12$ let me introduce a correction term $\epsilon(x)=x/12$ with the property $$ \epsilon(x)-\epsilon(x-1)=\frac1{12}.\tag{3} $$ Let's then define the function $$ F(x)=2(g(x)+\epsilon(x))=\frac13x^3+\frac12x^2+\frac16x. $$ Taking into account equations $(2)$ and $(3)$ we get the desired identity $$ F(x)-F(x-1)=x^2.\tag{4} $$ This telescopes the same way your telescoping with $f(x)$ did. Observe that $F(0)=0$. Of course, we have the expected factorization $$ F(x)=\frac16x(2x+1)(x+1). $$

It is hopefully clear how to get a formula for the sum of cubes by calculating appropriate integrals of $F(x)$, and produce a function $G(x)$ with the properties $G(0)=0$ and $$ G(x)-G(x-1)=x^3. $$ Rinse and repeat to your heart's content.

Jyrki Lahtonen
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