Q.Prove, by method of induction, for all $n \in N$: $1^2 + 4^2 + 7^2 + \ldots + (3n-2)^2 = \dfrac{n}{2}(6n^2-3n -1)$.

This is the well-known textbook induction result that $7^{2n}+2^{3n-3}\cdot 3^{n-1}$ is always divisible by 25 for every natural number $n$.

Check the base case $n=1$: $7^2 + 2^{0}\cdot 3^{0} = 49+1 = 50 = 25\times 2$. Divisible by 25.

Check $n=2$: $7^4 + 2^{3}\cdot 3^{1} = 2401 + 8\times 3 = 2401+24 = 2425 = 25\times 97$. Divisible by 25.

This pattern is proved in general by mathematical induction: assuming it holds for $n=k$ (i.e. $7^{2k}+2^{3k-3}\cdot 3^{k-1}=25m$ for some integer $m$), one shows $7^{2(k+1)}+2^{3(k+1)-3}\cdot 3^{k} = 49\cdot 7^{2k} + 24\cdot 2^{3k-3}\cdot 3^{k-1}$, which can be written as $49(7^{2k}+2^{3k-3}3^{k-1}) - 25\cdot 2^{3k-3}3^{k-1}$ — both terms are multiples of 25, completing the induction step.