Question

The sum of the following series \(1+6+\frac{9\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}} \right)}{7}+\frac{12\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}} \right)}{9}+\frac{15\left( {{1}^{2}}+{{2}^{2}}+\ldots +{{5}^{2}} \right)}{11}+\cdots\) up to 15 terms is:
1. 7520
2. 7510
3. 7830
4. 7820

Answer 1

Correct Answer - Option 4 : 7820

From question, the equation given is:

\(\Rightarrow S=1+6+\frac{9\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}} \right)}{7}+\frac{12\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}} \right)}{9}+\frac{15\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+{{5}^{2}} \right)}{11}+\ldots {{n}^{2}}\) 

\(\Rightarrow S=\frac{3\cdot {{(1)}^{2}}}{3}+\frac{6\cdot \left( {{1}^{2}}+{{2}^{2}} \right)}{5}+\frac{9\cdot \left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}} \right)}{7}+\frac{12\cdot \left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}} \right)}{9}+\ldots ..{{\text{n}}^{2}}\) 

The general form of the given series is written as:

\(\Rightarrow {{t}_{n}}=\frac{3n\cdot \left( {{1}^{2}}+{{2}^{2}}+\ldots .+{{n}^{2}} \right)}{\left( 2n+1 \right)}\) 

\(\Rightarrow {{t}_{n}}=\frac{3n\cdot n\left( n+1 \right)\left( 2n+1 \right)}{6\left( 2n+1 \right)}\) 

\(\Rightarrow {{t}_{n}}=\frac{{{n}^{3}}+{{n}^{2}}}{2}\) 

\(\therefore {{t}_{n}}=\frac{1}{2}\left[ {{n}^{3}}+{{n}^{2}} \right]\) 

\(\because \left[ \begin{matrix} \underset{k=1}{\overset{n}{\mathop \sum }}\,{{k}^{2}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \\ \underset{k=1}{\overset{n}{\mathop \sum }}\,{{k}^{3}}={{\left( \frac{n\left( n+1 \right)}{2} \right)}^{2}} \\ \end{matrix} \right]\) 

\(\Rightarrow {{S}_{n}}=\Sigma {{t}_{n}}=\frac{1}{2}\left\{ {{\left( \frac{n\left( n+1 \right)}{2} \right)}^{2}}+\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right\}\) 

\(\Rightarrow {{S}_{n}}=\frac{n\left( n+1 \right)}{4}\left( \frac{n\left( n+1 \right)}{2}+\frac{2n+1}{3} \right)\) 

\(\Rightarrow {{S}_{n}}=\frac{n\left( n+1 \right)}{4}\left( \frac{{{n}^{2}}+n}{2}+\frac{2n+1}{3} \right)\) 

\(\Rightarrow {{S}_{n}}=\frac{n\left( n+1 \right)}{4}\left( \frac{3{{n}^{2}}+3n+4n+2}{6} \right)\) 

\(\Rightarrow {{S}_{n}}=\frac{n\left( n+1 \right)}{4}\left( \frac{3{{n}^{2}}+7n+2}{6} \right)\) 

From question, we need to find sum of the series up to 15 terms. So, n =15,

\(\Rightarrow {{S}_{15}}=\frac{15\left( 15+1 \right)}{4}\left( \frac{\left( 3\times {{15}^{2}} \right)+\left( 7\times 15 \right)+2}{6} \right)\) 

\(\Rightarrow {{S}_{15}}=\frac{15\left( 16 \right)}{4}\left( \frac{\left( 3\times 225 \right)+\left( 105 \right)+2}{6} \right)\) 

\(\Rightarrow {{S}_{15}}=60\left( \frac{675+105+2}{6} \right)\) 

⇒ S₁₅ = 10(675 + 105 + 2)

⇒ S₁₅ = 10(782)

∴ S₁₅ = 7820