The sum $$\frac{1}{1 + 1^2 + 1^4} + \frac{2}{1 + 2^2 + 2^4} + \frac{3}{1 + 3^2 + 3^4} + \frac{3}{1 + 3^2 + 3^4} + ....... + \frac{99}{1 + 99^2 + 99^4}$$ lies between
Solution
$$\frac{1}{1 + 1^2 + 1^4} + \frac{2}{1 + 2^2 + 2^4} + \frac{3}{1 + 3^2 + 3^4}+ ....... + \frac{99}{1 + 99^2 + 99^4}$$
= $$\frac{1}{3} + \frac{2}{21} + \frac{3}{91} + ....... +$$
= $$\frac{1}{1 \cdot 3} + \frac{2}{3 \cdot 7} + \frac{3}{7 \cdot 13} + ....... +$$
$$\frac{1}{1 \cdot 3}$$ this can be written as $$\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}\right)$$
$$ \frac{2}{3 \cdot 7}$$ this can be written as $$\frac{1}{2}\left(\frac{1}{3}-\frac{1}{7}\right)$$
$$\frac{3}{7 \cdot 13}$$ this can be written as $$\frac{1}{2}\left(\frac{1}{7}-\frac{1}{13}\right)$$
The whole expression can be written as
=$$\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}\right)$$+ $$\frac{1}{2}\left(\frac{1}{3}-\frac{1}{7}\right)$$+$$\frac{1}{2}\left(\frac{1}{7}-\frac{1}{13}\right)$$ +....
The terms get cancelled out and we will be remained with the first from the first term and the second from the last term.
The second term in each of these terms form a pattern.
$$(n+1)^2-(n+1)+1$$
For the first term n=1, $$(n+1)^2-(n+1)+1$$ = 3
For the second term n=2, $$(n+1)^2-(n+1)+1$$ = 7
For the second term n=3, $$(n+1)^2-(n+1)+1$$ = 13
For the 99th term n=99, $$(n+1)^2-(n+1)+1$$ = 9901.
The final value of the expression is $$\frac{1}{2}-\frac{1}{9901}$$
This will be in between 0.49 and 0.5
