Question

The sum
(1+1^2+1^4)} + {2/ (1+2^2+2^4)} + {3/ (1+3^2+3^4)} + {4/ (1+4^2+4^4)} +...{99/ (1+99^2+99^4)

Answer 1

Answer:

$\frac{4950}{9901}$

Step-by-step explanation:

Given :

To find the sum of :$\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+ \frac{3}{1+3^2+3^4}+...+\frac{99}{1+99^2+99^4}$

Solution :

$\frac{1}{1+1^2+1^4} + \frac{2}{1+2^2+2^4} + \frac{3}{1+3^2+3^4}+...+\frac{99}{1+99^2+99^4}$

99

∑   $\frac{i}{1+i^2+i^4}$

i=1

99

∑   $\frac{i}{(i^2+1)^2 - i^2}$

i=1

99

∑   $\frac{i}{(i^2 + i + 1)(i^2 - i + 1)}$

i=1

99

∑   $\frac{1}{2}[\frac{(i^2+i+1) - (i^2-i+1)}{(i^2+i+1)(i^2-i+1)}]$

i=1

 99

$\frac{1}{2}$ ∑   $(\frac{1}{(i^2-i+1)}-\frac{1}{(i^2+i+1)})$

 i=1

                             

$\frac{1}{2} (\frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{7} + \frac{1}{7} - ....+ \frac{1}{1+99^2+99^4})$

$\frac{1}{2}(\frac{1}{1} - \frac{1}{9901})$

$\frac{1}{2}( \frac{9900}{9901})$

$\frac{4950}{9901}$