Question

What is the sum of the series \[\dfrac{3}{4} + \dfrac{5}{{36}} + \dfrac{7}{{144}} + ...\] up to \[11\] terms?
A. \[\dfrac{{120}}{{121}}\]
B. \[\dfrac{{143}}{{144}}\]
C. 1
D. \[\dfrac{{144}}{{143}}\]

Answer 1

Hint: First, using the terms of the series find the general equation of the \[{n^{th}}\] term. Simplify the equation of the \[{n^{th}}\] term. After that, calculate the first \[11\] terms of the series using the \[{n^{th}}\] term and add them to get the required answer.

Formula Used:
\[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]

Complete step by step solution:
The given series is \[\dfrac{3}{4} + \dfrac{5}{{36}} + \dfrac{7}{{144}} + ...\].
Let’s calculate the general equation of the \[{n^{th}}\] term of the given series.
Clearly, each term of the given series is in the form \[\dfrac{{3 + 2\left( {n - 1} \right)}}{{{{\left( {n\left( {n + 1} \right)} \right)}^2}}}\].
So, the \[{n^{th}}\] term of the series is \[{a_n} = \dfrac{{3 + 2\left( {n - 1} \right)}}{{{{\left( {n\left( {n + 1} \right)} \right)}^2}}}\].
Let’s simplify the general term.
\[{a_n} = \dfrac{{3 + 2n - 2}}{{\left( {{n^2}{{\left( {n + 1} \right)}^2}} \right)}}\]
\[ \Rightarrow \]\[{a_n} = \dfrac{{2n + 1}}{{\left( {{n^2}{{\left( {n + 1} \right)}^2}} \right)}}\]
\[ \Rightarrow \]\[{a_n} = \dfrac{{{{\left( {n + 1} \right)}^2} - {n^2}}}{{\left( {{n^2}{{\left( {n + 1} \right)}^2}} \right)}}\]
Simplify the above term.
\[{a_n} = \dfrac{{{{\left( {n + 1} \right)}^2}}}{{\left( {{n^2}{{\left( {n + 1} \right)}^2}} \right)}} - \dfrac{{{n^2}}}{{\left( {{n^2}{{\left( {n + 1} \right)}^2}} \right)}}\]
\[ \Rightarrow \]\[{a_n} = \dfrac{1}{{{n^2}}} - \dfrac{1}{{{{\left( {n + 1} \right)}^2}}}\]

Let consider \[{S_{11}}\] be the sum of the first \[11\] terms of the series.
Use the equation of the \[{n^{th}}\] term.
\[{S_{11}} = \left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{{\left( {1 + 1} \right)}^2}}}} \right) + \left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{{\left( {2 + 1} \right)}^2}}}} \right) + \left( {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{{\left( {3 + 1} \right)}^2}}}} \right) + ... + \left( {\dfrac{1}{{1{1^2}}} - \dfrac{1}{{{{\left( {11 + 1} \right)}^2}}}} \right)\]
\[ \Rightarrow \]\[{S_{11}} = \left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right) + \left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right) + \left( {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{4^2}}}} \right) + ... + \left( {\dfrac{1}{{1{1^2}}} - \dfrac{1}{{1{2^2}}}} \right)\]
\[ \Rightarrow \]\[{S_{11}} = \dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^2}}} - \dfrac{1}{{{4^2}}} + ... + \dfrac{1}{{1{1^2}}} - \dfrac{1}{{1{2^2}}}\]
Cancel out the terms with opposite signs.
\[{S_{11}} = \dfrac{1}{{{1^2}}} - \dfrac{1}{{1{2^2}}}\]
\[ \Rightarrow \]\[{S_{11}} = 1 - \dfrac{1}{{144}}\]
\[ \Rightarrow \]\[{S_{11}} = \dfrac{{144 - 1}}{{144}}\]
\[ \Rightarrow \]\[{S_{11}} = \dfrac{{143}}{{144}}\]
Hence the correct option is B.

Note: A list of numbers in a specific order or pattern is called a sequence. The sum of the terms in a sequence is called a series.
The sum of \[n\] terms in any series is the addition of the first \[n\] terms in that series.