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.

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}
