Question

If $S_1,S_2,S_3$ are the sum of first $n$ natural numbers, their squares and their cubes, respectively, show that ${9S}_2^2=S_3(1+{8S}_1).$

Answer 1

From the given information
$S_1=\frac{n(n+1)}{2}{S}_3=\frac{n^2{(n+1)}^2}{4}\text{Here},S_3(1+{8S}_1)=\frac{n^2{(n+1)}^2}{4}[1+\frac{8n(n+1)}{2}]=\frac{n^2{(n+1)}^2}{4}[1+{4n}^2+4n]=\frac{n^2{(n+1)}^2}{4}{(2n+1)}^2=\frac{{\left[n\right(n+1\left)\right(2n+1\left)\right]}^2}{4}.....\left(1\right)\text{Also},9S_2^2=9\frac{{\left[n\right(n+1\left)\right(2n+1\left)\right]}^2}{6^2}=\frac{9}{36}{\left[n\right(n+1\left)\right(2n+1\left)\right]}^2=\frac{{\left[n\right(n+1\left)\right(2n+1\left)\right]}^2}{4}...\left(2\right)$
Thus, from (1) and (2) we obtain $9S_2^2=S_3(1+{8S}_1)$