Question

Sum to n terms of the given series

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

Answer 1

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

Given series is

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

Now, the rᵗʰ term of series is given by

$\rm \: T_r = \dfrac{r}{1 + {r}^{2} + {r}^{4} }$

can be rewritten as

$\rm \: = \: \dfrac{r}{ {r}^{4} + {r}^{2} + 1 }$

$\rm \: = \: \dfrac{r}{ {r}^{4} + {r}^{2} + {r}^{2} + 1 - {r}^{2} }$

$\rm \: = \: \dfrac{r}{ {r}^{4} + 2{r}^{2} + 1 - {r}^{2} }$

$\rm \: = \: \dfrac{r}{ {( {r}^{2} + 1)}^{2} - {r}^{2} }$

$\rm \: = \: \dfrac{r}{( {r}^{2} + 1 + r)( {r}^{2} + 1 - r)}$

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

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

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

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

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

So,

$\rm\implies \:\rm T_r\: = \: \dfrac{1}{2} \bigg(\dfrac{1}{ {r}^{2} + 1 - r} - \dfrac{1}{ {r}^{2} + 1 + r} \bigg)$

On substituting r = 1, 2, 3, ----, n, we get

$\rm T_1\: = \: \dfrac{1}{2} \bigg(\dfrac{1}{ 1} - \dfrac{1}{ 3} \bigg)$

$\rm T_2\: = \: \dfrac{1}{2} \bigg(\dfrac{1}{3} - \dfrac{1}{7} \bigg)$

$\rm T_3\: = \: \dfrac{1}{2} \bigg(\dfrac{1}{7} - \dfrac{1}{13} \bigg)$

.

.

.

.

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

Now, We know, Sum of n terms is

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

$\rm \: S_n = \dfrac{1}{2} \bigg(\dfrac{1}{ 1} - \dfrac{1}{ 3} \bigg) + \dfrac{1}{2} \bigg(\dfrac{1}{3} - \dfrac{1}{7} \bigg) + \dfrac{1}{2} \bigg(\dfrac{1}{7} - \dfrac{1}{ 13} \bigg) + - - + \dfrac{1}{2} \bigg(\dfrac{1}{ {n}^{2} + 1 - n} - \dfrac{1}{ {n}^{2} + 1 + n} \bigg)$

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

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

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

Hence,

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

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

ADDITIONAL INFORMATION

$\boxed{\tt{ 1 + 2 + 3 + - - - + n = \frac{n(n + 1)}{2}}}$

$\boxed{\tt{ {1}^{2} + {2}^{2} + {3}^{2} + - - - + {n}^{2} = \frac{n(n + 1)(2n + 1)}{6}}}$

$\boxed{\tt{ {1}^{3}+{2}^{3}+ {3}^{3} + - - - + {n}^{3} = {\bigg[\dfrac{n(n + 1)}{2} \bigg]}^{2} }}$

Answer 2

Question:-

Sum to n terms of the given series

$\sf \large \frac{1}{1 + {1}^{2} + {1}^{4} } + \frac{2}{1 + {2}^{2} + {2}^{4} } + \frac{3}{1 + {3}^{2} + {3}^{4} } + ...$

Given:-

$\sf \large \frac{1}{1 + {1}^{2} + {1}^{4} } + \frac{2}{1 + {2}^{2} + {2}^{4} } + \frac{3}{1 + {3}^{2} + {3}^{4} } + ...$

Solution:-

Now,

$\sf \large \: T_r = \frac{r}{ 1 + r {}^{2} + r {}^{4} } = \frac{r}{(1 + r {}^{4}) + r {}^{2} } = \frac{r}{(1 + r {}^{2} ) {}^{2} - r {}^{2} }$

$\sf \large = \frac{r}{(r {}^{2} + r + 1)(r {}^{2} - r + 1) }$

$\sf \large = \frac{1}{2} \times \frac{(r {}^{2} + r + 1) - (r {}^{2} - r + 1) }{(r {}^{2} + r + 1)(r {}^{2} - r +1) }$

$\sf \large = \frac{1}{2} \bigg \{ \frac{1}{r {}^{2} - r + 1} - \frac{1}{r {}^{2} + r + 1} \bigg \}$

$\sf \large There, Sum \: to \: n \: terms, \sf \large \: S_n = \sf{ \sum \limits_{r = 1}^{n} } \: T_r$

$\sf \large =\sf{ \sum \limits_{r = 1}^{n}} \: \frac{1}{2} \bigg ( \frac{1}{r {}^{2} - r + 1} - \frac{1}{r {}^{2} + r + 1 } \bigg)$

$\sf \large = \frac{1}{2} \bigg \{( \frac{1}{1} - {{ \cancel{\frac{1}{3} }}}) + ({{ \cancel{ \frac{1}{3}}}} - {{ \cancel{\frac{1}{7}}}} ) + ( {{ \cancel{\frac{1}{7} }}}- \frac{1}{13} ) + ...( \frac{1}{n {}^{2} - n + 1 } - \frac{1}{n {}^{2} + n + 1} ) \bigg \}$

$\sf \large = \frac{1}{2} \bigg \{1 - \frac{1}{n {}^{2} + n + 1 } \bigg \}$

$\sf \large = \frac{1}{2} \bigg( \frac{n {}^{2} + n }{n {}^{2} + n + 1} \bigg )$

Answer:-

${ \boxed{ \sf \large \red{\sf \large \: S_n = \frac{1}{2} \bigg( \frac{n {}^{2} + n }{n {}^{2} + n + 1} \bigg )}}}$

Hope you have satisfied. ⚘