Question

The sum of first 9 terms of series $\dfrac{{{1}^{3}}}{1}+\dfrac{{{1}^{3}}+{{2}^{3}}}{1+3}+\dfrac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}{1+3+5}+..........$ is
(a) 71
(b) 96
(c) 142
(d) 192

Answer 1

Hint: What we will do first is, we will find the pattern in numerator and denominator of the series and generalise the term of series as ${{T}_{n}}$, then we will find the summation of pattern in numerator and denominator in terms of n and then solve for n = 11.

Complete step-by-step answer:
We are asked to find the sum of 9 terms of given series which is $\dfrac{{{1}^{3}}}{1}+\dfrac{{{1}^{3}}+{{2}^{3}}}{1+3}+\dfrac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}{1+3+5}+..........$.
Now, what we can see in the series, there is a pattern in both numerator and denominator.
Now, firstly if we see patterns in numerator then, we can see that there is a series of sum of cube of first n natural numbers.
So, we can denote numerator as $={{1}^{3}}+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+{{5}^{3}}+{{6}^{3}}.........+{{n}^{3}}$
Now, firstly if we see pattern in denominator then, we can see that there is arithmetic progression whose first term a = 1 , common difference = 2
So, we can denote numerator as $=\dfrac{n}{2}\left( 2(1)+(n-1)2 \right)$, as sum of first n terms of A.P whose first term is a and common difference is d is denoted as $\dfrac{n}{2}\left( 2a+(n-1)d \right)$.
So, as we have seen the pattern of given series $\dfrac{{{1}^{3}}}{1}+\dfrac{{{1}^{3}}+{{2}^{3}}}{1+3}+\dfrac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}{1+3+5}+..........$
Then, we can write the ${{n}^{th}}$ term of this series as,
${{T}_{n}}=\dfrac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+{{5}^{3}}+{{6}^{3}}.........+{{n}^{3}}}{\dfrac{n}{2}\left( 2(1)+(n-1)2 \right)}$ , where ${{T}_{n}}$ denotes ${{n}^{th}}$term of series $\dfrac{{{1}^{3}}}{1}+\dfrac{{{1}^{3}}+{{2}^{3}}}{1+3}+\dfrac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}{1+3+5}+...$
On simplifying, we get
${{T}_{n}}=\dfrac{{{\left( \dfrac{n(n+1)}{2} \right)}^{2}}}{\dfrac{n}{2}\left( 2(1)+(n-1)2 \right)}$, as sum of cube of first n natural number is ${{\left( \dfrac{n(n+1)}{2} \right)}^{2}}$.
Again on simplifying, we get
\[{{T}_{n}}=\dfrac{\dfrac{{{n}^{2}}{{(n+1)}^{2}}}{4}}{\dfrac{n}{2}\left( 2n \right)}\]
\[{{T}_{n}}=\dfrac{{{(n+1)}^{2}}}{4}\]
For, n = 1 we get \[{{T}_{1}}=\dfrac{{{(2)}^{2}}}{4}\]
For, n = 2 we get \[{{T}_{2}}=\dfrac{{{(3)}^{2}}}{4}\]
.
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Now, we have \[{{2}^{2}}+{{3}^{2}}+{{4}^{2}}......\] in numerator of \[{{T}_{n}}=\dfrac{{{(n+1)}^{2}}}{4}\] but \[{{(1)}^{2}}\]is missing so we can write numerator of \[{{T}_{n}}=\dfrac{{{(n+1)}^{2}}}{4}\]as \[\sum\limits_{n=1}^{n}{{{(n)}^{2}}}-1\] because \[\sum\limits_{n=1}^{n}{{{(n)}^{2}}}={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+......\] and \[\sum\limits_{n=1}^{n}{{{(n)}^{2}}}-1={{2}^{2}}+{{3}^{2}}+......\]
\[{{T}_{n}}=\dfrac{\sum\limits_{n=1}^{n}{{{(n)}^{2}}}-1}{4}\]
We know that \[\sum\limits_{n=1}^{n}{{{(n)}^{2}}}={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+......=\dfrac{n(n+1)(2n+1)}{6}\]
So, \[{{T}_{n}}=\dfrac{\dfrac{n(n+1)(2n+1)}{6}-1}{4}\]
At, n = 11
\[{{T}_{11}}=\dfrac{\dfrac{11(11+1)(2(11)+1)}{6}-1}{4}\]
On, solving, we get
\[{{T}_{11}}=\dfrac{384}{4}\]
\[{{T}_{11}}=96\]

So, the correct answer is “Option (b)”.

Note: To solve such a typical question, try to find a pattern in the numerator and denominator of the series first then substitute the formula of summation of that pattern and then solve for any value of n. Calculation is a priority for these types of questions so try to avoid calculation error.