Question

The sum of first 9 terms of the series $ \dfrac{{{1}^{3}}}{1}+\dfrac{{{1}^{3}}+{{2}^{3}}}{1+3}+\dfrac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}{1+3+5}+... $ \[\]
A.71\[\]
B.96\[\]
C. 142\[\]
D.192\[\]

Answer 1

Hint: We find the general term $ {{t}_{n}} $ of the given series by observing that the general term will have its numerator as sum of first cubed $ n $ numbers that is $ {{1}^{3}}+{{2}^{3}}+...+{{n}^{3}}={{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}} $ and denominator as sum of first $ n $ odd numbers that is $ 1+3+5....+\left( 2n-1 \right)={{n}^{2}} $ . We use a summation operator on $ {{t}_{n}} $ and simplify to get the value of the series up to $ n $ numbers. We put $ n=9 $ to get the required values.

Complete step-by-step answer:
We know that if $ {{t}_{1}},{{t}_{2}},{{t}_{3}},... $ is an infinite sequence then its series can be denoted in summation form as
\[\sum\limits_{n=1}^{\infty }{{{t}_{n}}}={{t}_{1}}+{{t}_{2}}+...\]
Here $ {{t}_{n}} $ is called $ {{n}^{\text{th}}} $ term or general term of the series and $ \Sigma $ is the summation operator. We are given the following series in the question.
\[\dfrac{{{1}^{3}}}{1}+\dfrac{{{1}^{3}}+{{2}^{3}}}{1+3}+\dfrac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}{1+3+5}+...\]
We have to first find the general term. We see that the numerator of the general term will be $ {{1}^{3}}+{{2}^{3}}+{{3}^{3}}...+{{n}^{3}} $ that is the sum of first cubed $ n $ numbers whose formula we know as $ \sum\limits_{n=1}^{n}{{{n}^{3}}}={{\left\{ \dfrac{n\left( n+1 \right)}{2} \right\}}^{2}} $ . The denominator of the general term will be $ 1+3+5+....+\left( 2n-1 \right) $ that is the sum of first $ n $ odd numbers whose formula we know $ \sum\limits_{n=1}^{n}{\left( 2n-1 \right)}={{n}^{2}} $ . So the general term is ;
\[\begin{align}
  & {{t}_{n}}=\dfrac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+...+{{n}^{3}}}{1+3+5+...+\left( 2n-1 \right)} \\
 & \Rightarrow {{t}_{n}}=\dfrac{{{\left( \dfrac{n(n+1)}{2} \right)}^{2}}}{{{n}^{2}}} \\
 & \Rightarrow {{t}_{n}}=\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4{{n}^{2}}}=\dfrac{{{\left( n+1 \right)}^{2}}}{4} \\
\end{align}\]
We use summation operators on the general term to have the value so the series up to $ {{n}^{\text{th}}} $ term. We have;
\[\begin{align}
  & \sum\limits_{n=1}^{n}{{{t}_{n}}}=\sum\limits_{n=1}^{n}{\dfrac{{{\left( n+1 \right)}^{2}}}{4}} \\
 & \Rightarrow \sum\limits_{n=1}^{n}{{{t}_{n}}}=\dfrac{1}{4}\sum\limits_{n=1}^{n}{{{\left( n+1 \right)}^{2}}} \\
\end{align}\]
We use the algebraic identity of $ {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} $ for $ a=n,b=1 $ in the above step to have;
\[\begin{align}
  & \Rightarrow \sum\limits_{n=1}^{n}{{{t}_{n}}}=\dfrac{1}{4}\sum\limits_{n=1}^{n}{\left( {{n}^{2}}+2n+1 \right)} \\
 & \Rightarrow \sum\limits_{n=1}^{n}{{{t}_{n}}}=\dfrac{1}{4}\left( \sum\limits_{n=1}^{n}{{{n}^{2}}+\sum\limits_{n=1}^{n}{2n+\sum\limits_{n=1}^{n}{1}}} \right) \\
 & \Rightarrow \sum\limits_{n=1}^{n}{{{t}_{n}}}=\dfrac{1}{4}\left( \sum\limits_{n=1}^{n}{{{n}^{2}}+2\sum\limits_{n=1}^{n}{n+\sum\limits_{n=1}^{n}{1}}} \right) \\
\end{align}\]
We know from the sum of first squared $ n $ numbers that $ \sum\limits_{n=1}^{n}{{{n}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} $ , the sum of first $ n $ numbers that $ \sum\limits_{n=1}^{n}{n}=\dfrac{n\left( n+1 \right)}{2} $ and if we add 1 $ n $ times we shall get $ \sum\limits_{n=1}^{n}{1}=n $ . So we have;
\[\Rightarrow \sum\limits_{n=1}^{n}{{{t}_{n}}}=\dfrac{1}{4}\left( \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+\dfrac{n\left( n+1 \right)}{2}+n \right)\]
The above obtained expression is the value of series up to $ {{n}^{\text{th}}} $ term. We put $ n=9 $ in the above step to get the sum first 9 terms as;
\[\begin{align}
  & \Rightarrow \sum\limits_{n=1}^{9}{{{t}_{n}}}=\dfrac{1}{4}\left( \dfrac{9\left( 9+1 \right)\left( 2\cdot 9+1 \right)}{6}+\dfrac{9\left( 9+1 \right)}{2}+9 \right) \\
 & \Rightarrow \sum\limits_{n=1}^{9}{{{t}_{n}}}=\dfrac{1}{4}\left( \dfrac{9\times 10\times 19}{6}+\dfrac{9\times 10}{2}+9 \right) \\
 & \Rightarrow \sum\limits_{n=1}^{9}{{{t}_{n}}}=\dfrac{1}{4}\left( 285+90+9 \right)=\dfrac{1}{4}\times 384=96 \\
\end{align}\]

So, the correct answer is “Option B”.

Note: The difference between series and sequence e is that in sequence terms are separated by a comma and in series the terms are separated by addition symbols. We have used distributive law o of summation $ \sum\limits_{n=a}^{b}{C\cdot {{t}_{n}}}=C\cdot \sum\limits_{n=a}^{b}{{{t}_{n}}} $ where $ C $ is constant and associative law $ \sum\limits_{n=a}^{b}{\left( f\left( n \right)+g\left( n \right) \right)}=\sum\limits_{n=a}^{b}{f\left( n \right)}+\sum\limits_{n=a}^{b}{g\left( n \right)} $