Theorem needed to prove a summation

Question

I know that the following relation holds:

$$\sum_{x=1}^y\frac{x(5x+6)}{45}=\frac{y(y+1)(10y+23)}{270}$$

But what theorem should I use to prove that relation?

Answer 1

It is known that $\sum_{x=1}^y x=\frac{y(y+1)}{2}$ and $\sum_{x=1}^y x^2=\frac{y(y+1)(2y+1)}{6}$. Refer to here for example.

Your required sum will become:

$$\sum_{x=1}^y \frac{x^2}{9} + \frac{2}{15}x$$

$$=\frac{1}{9}\cdot\frac{y(y+1)(2y+1)}{6}+\frac{2}{15}\cdot\frac{y(y+1)}{2}$$

$$=y(y+1)\left[\frac{2y+1}{54}+\frac{1}{15}\right]$$

$$=y(y+1)\left[\frac{10y+5}{270}+\frac{18}{270}\right]$$

$$=y(y+1)\left[\frac{10y+23}{270}\right]$$

$$=\frac{y(y+1)(10y+23)}{270}$$