The sum of the first $n$ squares $(1 + 4 + 9 + \cdots + n^2)$ is $\frac{n(n+1)(2n+1)}{6}$ Prove that the sum of the first $n$ squares $(1 + 4 + 9 + \cdots + n^2)$ is 
$\frac{n(n+1)(2n+1)}{6}$.
Can someone pls help and provide a solution for this and if possible explain the question 

Let it be true for k so for  $k^2=\frac{(k)(k+1)(2k+1)}{6}=$..(1). Now we will have to prove for $(k+1)^2$. So we have to prove $(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}$. So we start ${(1^2+2^2...k^2+(k+1)^2)}=\frac{(k)(k+1)(2k+1)}{6}.(k^2+2k+1)=\frac{(k+1)(2k^2+k+6k+6)}{6}=\frac{(k+1)(k+2)(2k+3)}{6}$. Hope now you know factorizations and also hope its clear.
