Question

If the sum of the ten terms of the series ${\left( {1\dfrac{3}{5}} \right)^2} + {\left( {2\dfrac{2}{5}} \right)^2} + {\left( {3\dfrac{1}{5}} \right)^2} + {4^2} + {\left( {4\dfrac{4}{5}} \right)^2} + ........$ is $\dfrac{{16}}{5}m$ , then $m$ is equal to:
(A) $102$
(B) $101$
(C) $100$
(D) $99$

Answer 1

Hint: In this question first we will convert mixed fraction the series into proper fraction series and then from that series we will find the equation for the general term of this series and then we will find the sum of first ten terms of this series with the help of some formulas.

Complete step-by-step answer:
The given series is ${\left( {1\dfrac{3}{5}} \right)^2} + {\left( {2\dfrac{2}{5}} \right)^2} + {\left( {3\dfrac{1}{5}} \right)^2} + {4^2} + {\left( {4\dfrac{4}{5}} \right)^2} + ........$
Now, we know that $a\dfrac{c}{d} = \dfrac{{a \times d + c}}{d}$ . Therefore, we can write the above series as:
$
   \Rightarrow {\left( {\dfrac{8}{5}} \right)^2} + {\left( {\dfrac{{12}}{5}} \right)^2} + {\left( {\dfrac{{16}}{5}} \right)^2} + {\left( {\dfrac{{20}}{5}} \right)^2} + {\left( {\dfrac{{24}}{5}} \right)^2} + ........ \\
   \Rightarrow \dfrac{{{8^2} + {{12}^2} + {{16}^2} + {{20}^2} + {{24}^2} + ......}}{{25}}\,\,\,\_\_\_\_\_\_\left( 1 \right) \\
 $
Now, look at numerator of the above series and find its ${n^{th}}$ term. The ${n^{th}}$ term of the numerator is ${T_n} = {\left[ {4\left( {n + 1} \right)} \right]^2}$ where $n \in 1\,{\text{to}}\,10$ as there are ten terms in the series. Therefore, we can write ${T_n} = 16\left( {{n^2} + 2n + 1} \right)$
Now, we will find the sum of the series
$
  {S_n} = \sum\limits_{n = 1}^{10} {16\left( {{n^2} + 2n + 1} \right)} \\
   \Rightarrow {S_n} = 16\left[ {\sum\limits_{n = 1}^{10} {{n^2}} + \sum\limits_{n = 1}^{10} {2n} + \sum\limits_{n = 1}^{10} 1 } \right] \\
 $
Now, we will use the formula of sum of squares of $n$ natural numbers and formula of sum of $n$ natural numbers to find the sum of above series.
\[ \Rightarrow {S_n} = 16\left[ {\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + 2\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right) + n} \right]\]
Now, put $n = 10$ in the above equation. Therefore, we will get:
$
   \Rightarrow {S_n} = 16\left[ {\dfrac{{10\left( {10 + 1} \right)\left( {2\left( {10} \right) + 1} \right)}}{6} + 2\left( {\dfrac{{10\left( {10 + 1} \right)}}{2}} \right) + 10} \right] \\
   \Rightarrow {S_n} = 16\left[ {\dfrac{{10\left( {11} \right)\left( {21} \right)}}{6} + 2\left( {\dfrac{{10\left( {11} \right)}}{2}} \right) + 10} \right] \\
 $
Now, simplify the above equation
$ \Rightarrow {S_n} = 16\left[ {385 + 110 + 10} \right] = 16\left[ {505} \right]$
Therefore, substitute the above value in the numerator of the equation $\left( 1 \right)$
$ \Rightarrow \dfrac{{16\left[ {505} \right]}}{{25}} = \dfrac{{16\left( {101} \right)}}{5}$
Now, compare $\dfrac{{16\left( {101} \right)}}{5}$ with $\dfrac{{16}}{5}m$
Hence, the value of $m$ is $101$ .
Hence, the correct option is (B).

Note: In this question the important thing is the conversion of the mixed fraction series into proper fraction series because after conversion we will be able to find the ${n^{th}}$ term of the series. The other important thing is that we should be able to recall the formula of sum of squares of $n$ natural numbers and the formula of sum of $n$ natural numbers.