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 
 A: *

*check if the statement holds true for $n=1$:
$$1^2=1=\frac{1(1+1)(2\cdot 1+1)}{6}=\frac{6}{6}=1$$

*Inductive step: show that if statement holds for $k$, then it also holds for $k+1$. This is done as follows:
$$1^2+2^2+3^2+\ldots+k^2=\frac{k(k+1)(2k+1)}{6}$$
$$1^2+2^2+3^2+\ldots+k^2+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$$
Note that the last equation may be rewritten as:
$$\frac{k(k+1)(2k+1)}{6}+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$$
It remains that both sides are indeed the same.

A: By mathematical induction, say $P(n):1+4+9+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ 
Now $P(1)$ is true as L.H.S. $= 1$ and R.H.S. $= 1$
Say $P(k)$ is true for some $k \in \mathbb{N}$ and $k>1$.
Therefore $1+4+9+\cdots+k^2=\frac{k(k+1)(2k+1)}{6}$
Now for $P(k+1)$,
\begin{align}
& 1+4+9+\cdots+k^2+(k+1)^2 \\[10pt]
= {} & \frac{k(k+1)(2k+1)}{6}+(k+1)^2 \\[10pt]
= {} & (k+1)\cdot \frac{2k^2+k+6(k+1)}{6} \\[10pt]
= {} &\frac{(k+1)[k+1+1][2(k+1)+1]}{6}
\end{align}
So $P(k+1)$ is true.
Hence by induction, $P(n):1+4+9+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ is true.
A: 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.
