# (c) Prove that 12 +22+32+ …+ n2 = n(n+1)(2n+1)/6 ;  n ∈ N

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The given expression is a sum of squares of consecutive natural numbers. We can prove the formula for the sum of squares by mathematical induction.

Base Case (n = 1):

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

The formula holds for n = 1.

Inductive Step:

Assume that the formula holds for some arbitrary positive integer k, i.e.,

$1^2 + 2^2 + 3^2 + \ldots + k^2 = \frac{k(k+1)(2k+1)}{6}.$

Now, let's prove it for (k+1):

$1^2 + 2^2 + 3^2 + \ldots + k^2 + (k+1)^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2.$

We can rewrite the denominator to have a common denominator:

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

Factor out (k+1) from the numerator:

$= \frac{(k+1)[k(2k+1) + 6(k+1)]}{6}.$

Simplify the expression in the brackets:

$= \frac{(k+1)[2k^2 + 7k + 6]}{6}.$

Factor the quadratic expression in the brackets:

$= \frac{(k+1)(k+2)(2k+3)}{6}.$

This matches the form of the formula for n = k+1:

$\frac{(k+1)[(k+1)+1][2(k+1)+1]}{6}.$

So, by mathematical induction, the formula holds for all positive integers n.