Question

If $S_1,S_2,S_3$ are the sums of first $n$ natural numbers, their squares, their cubes, respectively, then $\frac{S_3(1+8S_1)}{S_2^2}$ is equal to

• A. $1$
• B. $3$
• C. $9$
• D. $10$

Answer 1

The correct option is C $9$

We have formulas to calculate the sum of first $n$ natural numbers , their squares and also their cubes.
So, $S_1=$ Sum of first $n$ natural numbers $=\frac{n(n+1)}{2}$
$S_2=$ Sum of squares of first $n$ natural numbers $=\frac{n(n+1)(2n+1)}{6}$
$S_3=$ Sum of cubes of first $n$ natural numbers $={\left(\frac{n(n+1)}{2}\right)}^2$
So, $\frac{S_3(1+8S_1)}{S_2^2}=\frac{{\left(\frac{n(n+1)}{2}\right)}^2(1+8(\frac{n(n+1)}{2}\left)\right)}{(\frac{n(n+1)(2n+1)}{6}{)}^2}$
On simplifying, we get
$\frac{S_3(1+8S_1)}{S_2^2}=\frac{9(4n^2+4n+1)}{(2n+1{)}^2}=\frac{9(2n+1{)}^2}{(2n+1{)}^2}=9$