1. ## simplifying a statement

$(a(a+1) (2a + 1) )/ (6 ) + (a+1)^2 = ([a+1]([a+1]+1) (2[a+1]+1) )/ (6)$

also like this

a(a+1) (2a + 1) ------------- ---[a+1]([a+1]+1) (2[a+1]+1)
----------------+ (a+1)^2 =--------------------------
--------6 -------------------------------- 6

2. Hello, guenchy!

This seems to be an Inducation problem.

Something like . . .
. . Prove by induction that: . $1^2 + 2^2 + 3^2 + \cdots n^2 \;=\;\frac{n(n+1)(2n+1)}{6}$

$\frac{a(a+1) (2a + 1)}{6} + (a+1)^2 \;= \;\frac{(a+1)([a+1]+1) (2[a+1]+1)}{6} \;=$ . ${\color{blue}\frac{(a+1)(a+2)(2a+3)}{6}}$
You've verified $S(1)$.

You assumed $S(a)\!:\;1^2+ 2^2 + 3^2 + a^2 \;=\;\frac{a(a+1)(2a+1)}{6}$

You added $(a+1)^2$ to both sides:
. . $1^2 + 2^2 + 3^2 + \cdots + a^2 + (a+1)^2 \;=\;\frac{a(a+1)(2a+1)}{6} + (a+1)^2$

Now we want to show that the right side is the right side of $S(a+1)$,

. . which is: . $\frac{(a+1)([a+1]+1)(2[a+1]+1)}{6} \;=\;\frac{(a+1)(a+2)(2a+3)}{6}$

Simplify the left side . . .
It's just algebra, isn't it?

. . $\frac{a(a+1)(2a+1)}{6} + (a+1)^2 \;=\;\frac{a(a+1)(2a+1) + 6(a+1)^2}{6}$

Factor: . $\frac{(a+1)\,[a(2a+1) + 6(a+1)]}{6} \;=\;\frac{(a+1)(2a^2 + a + 6a + 6)}{6}$

. . $= \;\frac{(a+1)(2a^2+7a+6)}{6} \;=\;\frac{(a+1)(a+2)(2a+3)}{6}$ . . . . There!