Prove that binomial coefficient equals to sum of arithmetic progression I have to prove that
$\sum^{n}_{r=1}r^2 + \sum^{n}_{r=1}r = 2$${n+2}\choose{3}$
So far the only thing I can come up with is
${n(n+1)(2n+1)}\over{6}$ $+ {n(n+1)\over{2}} =$ $2{{n+2}\choose{3}}$
${n(n+1)(2n+1)}\over{3}$ $+ {n(n+1)} =$ ${{n+2}\choose{3}}$
$n(n+1)($${(2n+1)}\over{3}$ $+ 1) =$ ${{n+2}\choose{3}}$
And from here on I am stumped. Can anyone direct me or give me a hint as to what I should do next?
 A: The left side is a function of $n$,
not $r$.
So it is
$\dfrac{n(n+1)(2n+1)}{6}+ \dfrac{n(n+1)}{2}
=2\binom{n+2}{3}
$.
This should be easier.
A: Going from the first line of what you came up with to your second, you multiplied the left side by $2$ but divided the right side by $2$. Multiplying that side by $2$ instead gives that you're trying to prove
$$\frac{n(n + 1)(2n + 1)}{3} + n(n+1) = 4\binom{n+2}{3} \tag{1}\label{eq1A}$$
You have the right idea of factoring on the left side. Doing that gives
$$\begin{equation}\begin{aligned}
\frac{n(n + 1)(2n + 1)}{3} + n(n+1) & = n(n + 1)\left(\frac{2n + 1}{3} + 1\right) \\
& = n(n + 1)\left(\frac{2n + 4}{3}\right) \\
& = \frac{2n(n + 1)(n + 2)}{3}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Using the standard definition of the binomial coefficient being $\binom{a}{b} = \frac{a!}{b!(a - b)!} = \frac{a(a-1)\ldots (a - b + 1)}{b!}$ gives the right side of \eqref{eq1A} to be
$$\begin{equation}\begin{aligned}
4\binom{n+2}{3} & = 4\left(\frac{(n+2)(n+1)n}{1(2)(3)}\right) \\
& = \frac{2n(n + 1)(n + 2)}{3}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Thus, the left side and right sides of \eqref{eq1A} match, showing it's always true.
