Question

Let $S_1=\sum _{K=1}^{n}K,S_2=\sum _{K=1}^n{K}^2$ and $S_3=\sum _{K=1}^n{K}^3$, then $\frac{S_1^4{S}_2^2-S_2^2{S}_3^2}{S_1^2+S_3^2}$ is equal to

• A. $4$
• B. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is D $0$

$S_1=\sum k=\frac{n(n+1)}{2},S_2=\sum k^2=\frac{n(n+1)(2n+1)}{6},S_3=\sum k^3=\frac{n^2(n+1{)}^2}{4}$

$\frac{S_1^4{S}_2^2-S_2^2{S}_3^2}{S_1^2+S_3^2}$ $=\frac{\frac{n^4(n+1{)}^4}{2^4}.\frac{n^2(n+1{)}^2(2n+1{)}^2}{36}-\frac{n^2(n+1{)}^2(2n+1{)}^2}{36}.\frac{n^4(n+1{)}^4}{4^2}}{S_1^2+S_3^2}$

$=0$

$\therefore$ option D is correct