Prove that binomial coefficient equals to sum of arithmetic progression

Question

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?

Answer 1

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.

Answer 2

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.