Question

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

Answer 1

First we have to deduce the given series in the form of a sequence that it follows in order to find the ${n^{th}}$ term of the series. Then, we simplify the terms to make it easier for us to add them. Then, we can find the sum of the terms with a much simpler footprint that would be easier to calculate for us and then find the sum up to the required number of terms.

Complete step by step answer:
Let the sum of the series be denoted as ${S_n}$.
So, given, ${S_n} = \dfrac{3}{4} + \dfrac{5}{{36}} + \dfrac{7}{{144}} + .....$
We can write it as,
${S_n} = \dfrac{{\left( {2.1 + 1} \right)}}{{\left( {{1^2}{{.2}^2}} \right)}} + \dfrac{{\left( {2.2 + 1} \right)}}{{\left( {{2^2}{{.3}^2}} \right)}} + \dfrac{{\left( {2.3 + 1} \right)}}{{\left( {{3^2}{{.4}^2}} \right)}} + ....$
Thus, by observing the pattern of the terms of the series we can conclude that, ${n^{th}}$ term, ${T_n} = \dfrac{{\left( {2.n + 1} \right)}}{{\left[ {{n^2}.{{\left( {n + 1} \right)}^2}} \right]}}$
Now, adding and subtracting ${n^2}$ in the numerator, we get,
$\Rightarrow {T_n} = \dfrac{{\left( {{n^2} + 2.n + 1 - {n^2}} \right)}}{{\left[ {{n^2}.{{\left( {n + 1} \right)}^2}} \right]}}$
Simplifying the expression, we get,
$\Rightarrow {T_n} = \dfrac{{\left[ {\left( {{n^2} + 2.n + {1^2}} \right) - {n^2}} \right]}}{{\left[ {{n^2}.{{\left( {n + 1} \right)}^2}} \right]}}$
We know, ${(a + b)^2} = {a^2} + 2ab + {b^2}$, using this algebraic identity, we get,
$\Rightarrow {T_n} = \dfrac{{\left[ {{{\left( {n + 1} \right)}^2} - {n^2}} \right]}}{{\left[ {{n^2}.{{\left( {n + 1} \right)}^2}} \right]}}$
Now, diving each term in the numerator by the denominator, we get,
$\Rightarrow {T_n} = \dfrac{{{{\left( {n + 1} \right)}^2}}}{{{n^2}.{{\left( {n + 1} \right)}^2}}} - \dfrac{{{n^2}}}{{{n^2}.{{\left( {n + 1} \right)}^2}}}$
$\Rightarrow {T_n} = \dfrac{1}{{{n^2}}} - \dfrac{1}{{{{\left( {n + 1} \right)}^2}}}$
Now, putting $n = 1,2,3,4......$up to $n$ terms, we get,
$\Rightarrow {T_1} = \dfrac{1}{{{1^2}}} - \dfrac{1}{{{{\left( {1 + 1} \right)}^2}}} = \dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}$
$\Rightarrow {T_2} = \dfrac{1}{{{2^2}}} - \dfrac{1}{{{{\left( {2 + 1} \right)}^2}}} = \dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}$
$\Rightarrow {T_3} = \dfrac{1}{{{3^2}}} - \dfrac{1}{{{{\left( {3 + 1} \right)}^2}}} = \dfrac{1}{{{3^2}}} - \dfrac{1}{{{4^2}}}$
$\Rightarrow {T_4} = \dfrac{1}{{{4^2}}} - \dfrac{1}{{{{\left( {4 + 1} \right)}^2}}} = \dfrac{1}{{{4^2}}} - \dfrac{1}{{{5^2}}}$ and so on.
Hence, ${T_n} = \dfrac{1}{{{n^2}}} - \dfrac{1}{{{{\left( {n + 1} \right)}^2}}}$
Now, adding all the terms up to $n$ terms, we get,
${S_n} = {T_1} + {T_2} + {T_3} + ...... + {T_n}$
$\Rightarrow {S_n} = \left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right) + \left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right) + \left( {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{4^2}}}} \right) + ..... + \left( {\dfrac{1}{{{n^2}}} - \dfrac{1}{{{{\left( {n + 1} \right)}^2}}}} \right)$
$\Rightarrow {S_n} = 1 - \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^2}}} - \dfrac{1}{{{4^2}}} + ..... + \dfrac{1}{{{n^2}}} - \dfrac{1}{{{{\left( {n + 1} \right)}^2}}}$
Here, we can clearly observe that, all the terms have their negative counterparts and gets cancelled, except, $1$ and $\dfrac{1}{{{{\left( {n + 1} \right)}^2}}}$.
So, we get,
$\Rightarrow {S_n} = 1 - \dfrac{1}{{{{\left( {n + 1} \right)}^2}}}$
As per the question, we are to find the sum up to ${11^{th}}$ term.
Therefore, ${S_{11}} = 1 - \dfrac{1}{{{{\left( {11 + 1} \right)}^2}}}$
$\Rightarrow {S_{11}} = 1 - \dfrac{1}{{{{\left( {12} \right)}^2}}}$
$\Rightarrow {S_{11}} = 1 - \dfrac{1}{{144}}$
Simplifying the equation, we get,
$\Rightarrow {S_{11}} = \dfrac{{144 - 1}}{{144}}$
$\Rightarrow {S_{11}} = \dfrac{{143}}{{144}}$
Therefore, the sum of the series up to ${11^{th}}$ term is $\dfrac{{143}}{{144}}$.

Note: The sum of series gives us the idea where the series may converge to. Deduction of the terms gives us the idea about a particular term, where it may be in the series. The sum of series is also used to know how many terms are to be used in order to reach a particular value. We must know the methodology for solving such types of questions.