Question

Find the value of following expression if ${{S}_{1}},{{S}_{2}},{{S}_{3}}$ are the sums of first n natural numbers, their squares and their cubes respectively:
 $\dfrac{{{S}_{3}}\left( 1+8{{S}_{1}} \right)}{S_{2}^{2}}$
A. 1
B. 3
C. 9
D. 10

Answer 1

Hint: Put the values of ${{S}_{1}},{{S}_{2}}$ and ${{S}_{3}}$ in the given expressions where, ${{S}_{1}}=\dfrac{n\left( n+1 \right)}{2},{{S}_{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ and ${{S}_{3}}={{\left[ \dfrac{n\left( n+1 \right)}{2} \right]}^{2}}$.

Complete step-by-step answer:
We know that,
${{S}_{1}}=\dfrac{n\left( n+1 \right)}{2}$
${{S}_{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$
${{S}_{3}}={{\left[ \dfrac{n\left( n+1 \right)}{2} \right]}^{2}}$
where n is the number of natural numbers.
Substituting the values of ${{S}_{1}},{{S}_{2}}$ and ${{S}_{3}}$ in the expression \[\dfrac{{{S}_{3}}\left( 1+8{{S}_{1}} \right)}{S_{2}^{2}}\], we get:
\[\begin{align}
  & \dfrac{{{\left[ \dfrac{\left( n+1 \right)n}{2} \right]}^{2}}\left\{ 1+\dfrac{8n\left( n+1 \right)}{2} \right\}}{{{\left[ \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right]}^{2}}} \\
 & =\dfrac{1}{4}\times \dfrac{36{{n}^{2}}{{\left( n+1 \right)}^{2}}\left( 4{{n}^{2}}+4n+1 \right)}{{{n}^{2}}{{\left( n+1 \right)}^{2}}{{\left( 2n+1 \right)}^{2}}} \\
\end{align}\]
$4{{n}^{2}}+4n+1$ can be written as ${{\left( 2n+1 \right)}^{2}}$ as \[\left( {{a}^{2}}+2ab+{{b}^{2}} \right)={{\left( a+b \right)}^{2}}\].
\[\begin{align}
  & \Rightarrow \dfrac{{{S}_{3}}\left( 1+8{{S}_{1}} \right)}{S_{2}^{2}}=\dfrac{9\left[ {{\left( n \right)}^{2}}{{\left( n+1 \right)}^{2}}{{\left( 2n+1 \right)}^{2}} \right]}{{{n}^{2}}{{\left( n+1 \right)}^{2}}{{\left( 2n+1 \right)}^{2}}} \\
 & =9 \\
\end{align}\]
Answer is (C) 9.

Note: This problem is just a formula based problem so we have to use the following formulas to get the solution.
 \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\].
We know that,
${{S}_{1}}=\dfrac{n\left( n+1 \right)}{2}$
${{S}_{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{2}$
${{S}_{3}}={{\left[ \dfrac{n\left( n+1 \right)}{2} \right]}^{2}}$
where n is the number of natural numbers.