Question

Evaluate the following to n terms

$\frac{1}{1 - 3. {1}^{2} + {1}^{4} } + \frac{2}{1 - 3. {2}^{2} + {2}^{4} } + \frac{3}{1 - 3. {3}^{2} + {3}^{4} } + ...$
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Answer 1

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

Given series is

$\rm \: \dfrac{1}{1 - 3. {1}^{2} + {1}^{4} } + \dfrac{2}{1 - 3. {2}^{2} + {2}^{4} } + \dfrac{3}{1 - 3. {3}^{2} + {3}^{4} } + ...$

can be rewritten as

$\rm \:S_n = \dfrac{1}{1 - 3. {1}^{2} + {1}^{4} } + \dfrac{2}{1 - 3. {2}^{2} + {2}^{4} } + \dfrac{3}{1 - 3. {3}^{2} + {3}^{4} } + ...$

can be further rewritten as

$\rm \:S_n =\displaystyle\sum_{k=1}^n\rm \dfrac{k}{1 - 3. {k}^{2} + {k}^{4} }$

$\rm \:S_n =\displaystyle\sum_{k=1}^n\rm \dfrac{k}{{k}^{4} - {3k}^{2} + 1 }$

$\rm \:S_n =\displaystyle\sum_{k=1}^n\rm \dfrac{k}{{k}^{4} - {2k}^{2} - {k}^{2} + 1 }$

$\rm \:S_n =\displaystyle\sum_{k=1}^n\rm \dfrac{k}{{k}^{4} - {2k}^{2} + 1 - {k}^{2} }$

$\rm \:S_n =\displaystyle\sum_{k=1}^n\rm \dfrac{k}{({k}^{4} - {2k}^{2} + 1) - {k}^{2} }$

$\rm \:S_n =\displaystyle\sum_{k=1}^n\rm \dfrac{k}{ {( {k}^{2} - 1)}^{2} - {k}^{2} }$

$\rm \:S_n =\displaystyle\sum_{k=1}^n\rm \dfrac{k}{( {k}^{2} + k - 1)( {k}^{2} -k - 1)}$

$\rm \:S_n = \frac{1}{2} \displaystyle\sum_{k=1}^n\rm \dfrac{2k}{( {k}^{2} + k - 1)( {k}^{2} -k - 1)}$

$\rm \:S_n = \frac{1}{2} \displaystyle\sum_{k=1}^n\rm \dfrac{k + k+{k}^{2}-{k}^{2} + 1 - 1}{( {k}^{2} + k - 1)( {k}^{2} -k - 1)}$

$\rm \:S_n = \frac{1}{2} \displaystyle\sum_{k=1}^n\rm \dfrac{( {k}^{2} + k - 1) - ( {k}^{2} - k - 1) }{( {k}^{2} + k - 1)( {k}^{2} -k - 1)}$

$\rm \:S_n = \frac{1}{2} \displaystyle\sum_{k=1}^n\rm \bigg(\dfrac{1}{ {k}^{2} - k - 1} - \dfrac{1}{ {k}^{2} + k - 1} \bigg)$

So, it means

$\rm \:T_k = \dfrac{1}{2} \bigg(\dfrac{1}{ {k}^{2} - k - 1} - \dfrac{1}{ {k}^{2} + k - 1} \bigg)$

On substituting k = 1, 2, 3, ...,n, we have

$\rm \:T_1 = \dfrac{1}{2} \bigg(\dfrac{1}{ - 1} - \dfrac{1}{1} \bigg)$

$\rm \:T_2 = \dfrac{1}{2} \bigg(\dfrac{1}{1} - \dfrac{1}{5} \bigg)$

$\rm \:T_3 = \dfrac{1}{2} \bigg(\dfrac{1}{5} - \dfrac{1}{11} \bigg)$

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$\rm \:T_n = \dfrac{1}{2} \bigg(\dfrac{1}{ {n}^{2} - n - 1} - \dfrac{1}{ {n}^{2} + n - 1} \bigg)$

So,

$\rm \: S_n = T_1 + T_2 + T_3 + \cdots + T_n$

$\rm \:S_n = \dfrac{1}{2} \bigg( - 1 - 1 + \dfrac{1}{1} - \dfrac{1}{5} + \frac{1}{5} - \frac{1}{11} + \cdots + \dfrac{1}{ {n}^{2} - n - 1} - \dfrac{1}{ {n}^{2} + n - 1} \bigg)$

$\rm \:S_n = \dfrac{1}{2} \bigg( - 1 - \dfrac{1}{ {n}^{2} + n - 1} \bigg)$

$\rm \:S_n = \dfrac{1}{2} \bigg(\dfrac{ - {n}^{2} - n + 1 - 1}{ {n}^{2} + n - 1} \bigg)$

$\rm \:S_n = \dfrac{1}{2} \bigg(\dfrac{ - {n}^{2} - n}{ {n}^{2} + n - 1} \bigg)$

$\rm \:S_n = \: - \: \dfrac{1}{2} \bigg(\dfrac{{n}^{2} + n}{ {n}^{2} + n - 1} \bigg)$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \:S_n = \: - \: \dfrac{1}{2} \bigg(\dfrac{{n}^{2} + n}{ {n}^{2} + n - 1} \bigg) \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

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

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

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

$\boxed{ \rm{ \:\displaystyle\sum_{k=1}^n\rm {k}^{3} = {\bigg[\dfrac{n(n + 1)}{2} \bigg]}^{2} \: }}$