<s>[INST] 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 
 [/INST] *

*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.

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