Q.If $S_1, S_2$ and $S_3$ are the sums of first $n$ natural numbers, their squares and their cubes respectively, then show that $9S_2^2=S_3(1+8S_1)$.

$S_1=\dfrac{n(n+1)}2, S_2=\dfrac{n(n+1)(2n+1)}6, S_3=\left[\dfrac{n(n+1)}2\right]^2=S_1^2$. RHS $=S_1^2(1+8S_1)=S_1^2+8S_1^3$. With $S_1=\dfrac{n(n+1)}2$: $S_1^2=\dfrac{n^2(n+1)^2}4$, $8S_1^3=n^3(n+1)^3$. So RHS $=\dfrac{n^2(n+1)^2}4\left[1+4n(n+1)\right]=\dfrac{n^2(n+1)^2(4n^2+4n+1)}4=\dfrac{n^2(n+1)^2(2n+1)^2}4$ (since $4n^2+4n+1=(2n+1)^2$). LHS $=9S_2^2=9\left[\ …

Substitute the standard closed forms for $S_1, S_2, S_3$ into both sides separately and simplify e …

Not spotting that $4n^2+4n+1$ factors as the perfect square $(2n+1)^2$, which is the key sim …