Question

Find the sum upto 10 terms of the series,
$\frac{3 \times {1}^{3} }{ {1}^{2} } + \frac{5 \times ( {1}^{3} + {2}^{3}) }{ {1}^{2} + {2}^{2} } + \frac{7 \times( {1}^{ 3} + {2}^{3} + {3}^{3} )}{ {1}^{2} + {2}^{2} + {3}^{2} } + ...$
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Answer 1

Given series,

${ \displaystyle \longrightarrow \frac{3 \times {1}^{3} }{ {1}^{2} } + \frac{5 \times ( {1}^{3} + {2}^{3}) }{ {1}^{2} + {2}^{2} } + \frac{7 \times( {1}^{ 3} + {2}^{3} + {3}^{3} )}{ {1}^{2} + {2}^{2} + {3}^{2} } + ...}$

Let's firstly find the general term of the given series.

The given series has a sequence of odd numbers in numerator, sum of cubes of natural numbers upto n, and sum of squares of n natural number is denominator. Therefore, the general term of the given series is given by,

${\implies T_n = \dfrac{(2n+1)(1^3+2^3+3^3+...+n^3)}{(1^2+2^2+3^2+...+n^2)}}$

We know that,

• $\displaystyle \sum\limits_{i=1}^n i^2=\dfrac{n(n+1)(2n+1)}{6}$

• $\displaystyle \sum\limits_{i=1}^n i^3 = \left(\dfrac{n(n+1)}{2}\right)^2$

Therefore,

${\implies T_n = \dfrac{(2n+1) \times \left(\dfrac{n(n + 1)}{2} \right)^{2} }{ \dfrac{n(n + 1)(2n + 1)}{6} }}$

${\implies T_n = \dfrac{\dfrac{ {n}^{2} (n + 1)^{2} }{2} }{ \dfrac{n(n + 1)}{3} }}$

$\implies T_n = \dfrac{ {n}(n + 1) }{2} \times 3$

$\boxed{\implies T_n = \dfrac{3 {n}^{2} }{2} + \frac{n}{2} }$

This is the general term of the given series.

Now, the sum of given series upto 10 terms is given by,

$\displaystyle\longrightarrow\sum\limits_{n=1}^{10}T_n$

$\displaystyle\longrightarrow\sum\limits_{n=1}^{10} \frac{3 {n}^{2} }{2} + \frac{3n}{2}$

$\displaystyle\longrightarrow \frac{3}{2} \left(\sum\limits_{n=1}^{10} {n}^{2} + n \right)$

$\displaystyle\longrightarrow \frac{3}{2} \left(\sum\limits_{n=1}^{10} {n}^{2} +\sum\limits_{n=1}^{10}n \right)$

${\displaystyle\longrightarrow \frac{3}{2} \left( \frac{10(10 + 1)(2(10) + 1)}{6} + \frac{(10)(10+ 1)}{2} \right)}$

${\displaystyle\longrightarrow \frac{3}{2} \left( \frac{10 \times 11 \times 21}{6} + \frac{10 \times 11}{2} \right)}$

${\displaystyle\longrightarrow \frac{3}{2} \left( 385+ 55 \right)}$

${\displaystyle\longrightarrow \frac{3}{2} \times 440}$

${\displaystyle\longrightarrow 3 \times 220}$

${\displaystyle\longrightarrow 66}0$

Hence the required sum of 660.