Question

The sum of the series {1(2/3)}²+{2(1)3)}²+3²+{3(2/3)}²+....to 10 term is

Answer 1

Step-by-step explanation:

Given Question

The sum of the series is

$\tt{ {\bigg[1\dfrac{2}{3} \bigg]}^{2} + {\bigg[2\dfrac{1}{3} \bigg]}^{2} + {3}^{2} + {\bigg[3\dfrac{2}{3} \bigg]}^{2} + - - 10 \: terms}$

$\green{\begin{gathered}\large{\sf{{\underline{Formula \: Used - }}}} \end{gathered}}$

$\boxed{\tt{ \displaystyle\sum_{k=1}^{n}1 = n \: }}$

$\boxed{\tt{ \displaystyle\sum_{k=1}^{n}k = \frac{n(n + 1)}{2} \: }}$

$\boxed{\tt{ \displaystyle\sum_{k=1}^{n} {k}^{2} = \frac{n(n + 1)(2n + 1)}{6} \: }}$

$\large\underline{\sf{Solution-}}$

Given series is

$\rm :\longmapsto\: {\bigg[1\dfrac{2}{3} \bigg]}^{2} + {\bigg[2\dfrac{1}{3} \bigg]}^{2} + {3}^{2} + {\bigg[3\dfrac{2}{3} \bigg]}^{2} + - - - 10 \: terms$

can be rewritten as

$\rm \: = \: {\bigg[\dfrac{5}{3} \bigg]}^{2} + {\bigg[\dfrac{7}{3} \bigg]}^{2} + {\bigg[\dfrac{9}{3} \bigg]}^{2} + {\bigg[\dfrac{11}{3} \bigg]}^{2} + - - - 10 \: terms$

$\rm \: = \: \dfrac{1}{9}[ {5}^{2} + {7}^{2} + {9}^{2} + - - - 10 \: terms \: ]$

Now, here, 5, 7, 9 forms an AP series with first term 5 and common difference 2.

So, its general term is given by 5 + ( n - 1 )2 = 5 + 2n - 2 = 2n + 3

So, above series can be represented as

$\rm \: = \: \dfrac{1}{9}\displaystyle\sum_{n=1}^{10}(2n + 3) ^{2}$

$\rm \: = \: \dfrac{1}{9}\displaystyle\sum_{n=1}^{10}\bigg[ {4n}^{2} + 9 + 12n\bigg]$

$\rm \: = \: \dfrac{1}{9}\bigg[\displaystyle\sum_{n=1}^{10} {4n}^{2} + \displaystyle\sum_{n=1}^{10}9 + 12\displaystyle\sum_{n=1}^{10}n\bigg]$

$\rm \: = \: \dfrac{1}{9}\bigg[4\displaystyle\sum_{n=1}^{10} {n}^{2} +9 \displaystyle\sum_{n=1}^{10}1 + 12\displaystyle\sum_{n=1}^{10}n\bigg]$

$\rm \: = \: \dfrac{4}{9}\bigg[\dfrac{10(10 + 1)(20 + 1)}{6} \bigg] + 10 + \dfrac{4}{3}\bigg[\dfrac{10(10 + 1)}{2} \bigg]$

$\rm \: = \: \dfrac{4}{9}\bigg[\dfrac{10(11)(21)}{6} \bigg] + 10 + \dfrac{4}{3}\bigg[\dfrac{10(11)}{2} \bigg]$

$\rm \: = \: \dfrac{1540}{9} + 10 + \dfrac{220}{3}$

$\rm \: = \: \dfrac{1540 + 90 + 660}{9}$

$\rm \: = \: \dfrac{2290}{9}$

Hence,

$\boxed{\tt{ {\bigg[1\dfrac{2}{3} \bigg]}^{2} + {\bigg[2\dfrac{1}{3} \bigg]}^{2} + {3}^{2} + {\bigg[3\dfrac{2}{3} \bigg]}^{2} + - - 10 \: terms = \frac{2290}{9}}}$