Direct telescopic proof for sum of $1^2+2^2+...+n^2$ 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.
 A: 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 $$a_n=2\binom{n+1}{3}+\binom{n+1}{2}=\frac13 n(n^2-1)+\frac12 n(n+1)=\frac16 n(n+1)(2n+1).$$
A: 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}
A: 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.
