Question

If $S_1,S_2,S_3$ are the sum of the first $n$ natural numbers, their squares, and 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

$9$

Finding the value of $\frac{S_3(1+8S_1)}{{S_2}^2}$:

Step 1: Solve using known results of sum of numbers

The sum of the $n$ natural numbers is given by

${\mathit{S}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{2}}\dots \left(1\right)$

The sum of the squares of the $n$ natural numbers is given as,

${\mathit{S}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{6}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\dots \left(2\right)$

The sum of the cubes of the $n$ natural numbers is given as,

${\mathit{S}}_{\mathbf{3}}\mathbf{=}{\mathbf{\left(}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{2}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\dots \left(3\right)$

Step 2: Calculation for the given expression

Use the values of the above three sums to determine the required,

$\begin{array}{rcl}\frac{S_3(1+8S_1)}{{S_2}^2}& =& \frac{{\left({\displaystyle \frac{n\left(n+1\right)}{2}}\right)}^2\left(1+8\left({\displaystyle \frac{n\left(n+1\right)}{2}}\right)\right)}{{\left({\displaystyle \frac{n\left(n+1\right)\left(2n+1\right)}{6}}\right)}^2}\\ & =& \frac{{\displaystyle \frac{{\left(n^2+n\right)}^2}{4}}\left(1+4\left(n^2+n\right)\right)}{{\displaystyle \frac{{\left(n^2+n\right)}^2{\left(2n+1\right)}^2}{36}}}\\ & =& \frac{{\left(n^2+n\right)}^2}{4}\left(1+4\left(n^2+n\right)\right)\times \frac{36}{{\left(n^2+n\right)}^2{\left(2n+1\right)}^2}\\ & =& \frac{9}{{\left(2n+1\right)}^2}\left(4n^2+4n+1\right)\\ & =& \frac{9}{4n^2+4n+1}\left(4n^2+4n+1\right)\\ & =& 9\end{array}$

Hence, the correct option is (C).