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Question: Find the value of following expression if \({{S}_{1}},{{S}_{2}},{{S}_{3}}\) are the sums of first n ...

Find the value of following expression if S1,S2,S3{{S}_{1}},{{S}_{2}},{{S}_{3}} are the sums of first n natural numbers, their squares and their cubes respectively:
S3(1+8S1)S22\dfrac{{{S}_{3}}\left( 1+8{{S}_{1}} \right)}{S_{2}^{2}}
A. 1
B. 3
C. 9
D. 10

Explanation

Solution

Hint: Put the values of S1,S2{{S}_{1}},{{S}_{2}} and S3{{S}_{3}} in the given expressions where, S1=n(n+1)2,S2=n(n+1)(2n+1)6{{S}_{1}}=\dfrac{n\left( n+1 \right)}{2},{{S}_{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} and S3=[n(n+1)2]2{{S}_{3}}={{\left[ \dfrac{n\left( n+1 \right)}{2} \right]}^{2}}.

Complete step-by-step answer:
We know that,
S1=n(n+1)2{{S}_{1}}=\dfrac{n\left( n+1 \right)}{2}
S2=n(n+1)(2n+1)6{{S}_{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}
S3=[n(n+1)2]2{{S}_{3}}={{\left[ \dfrac{n\left( n+1 \right)}{2} \right]}^{2}}
where n is the number of natural numbers.
Substituting the values of S1,S2{{S}_{1}},{{S}_{2}} and S3{{S}_{3}} in the expression S3(1+8S1)S22\dfrac{{{S}_{3}}\left( 1+8{{S}_{1}} \right)}{S_{2}^{2}}, we get:

& \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{aligned}$$ $4{{n}^{2}}+4n+1$ can be written as ${{\left( 2n+1 \right)}^{2}}$ as $$\left( {{a}^{2}}+2ab+{{b}^{2}} \right)={{\left( a+b \right)}^{2}}$$. $$\begin{aligned} & \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{aligned}$$ Answer is (C) 9. Note: This problem is just a formula based problem so we have to use the following formulas to get the solution. $${{\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.