Please help me out with this problem. $$\frac{1}{(1+1^2+1^4)} + \frac{2}{(1+2^2+2^4)} + \cdots+ \frac{99}{(1+99^2+99^4)}$$ lies between
$(A)$ $0.46$ and $0.47$.
$(B)$ $0.52$ and $1.0$.
$(C)$ $0.48$ and $0.49$.
$(D)$ $0.49$ and $0.50$.
Explain the procedure
By using partial fractions you get
\begin{align*} \frac{k}{1+k^2+k^4}&=\frac{\frac12}{k^2-k+1}-\frac{\frac12}{k^2+k+1}\\[4pt] &=\frac12\left[\frac{1}{\left(k-\frac12\right)^2+\frac34}-\frac{1}{\left(k+\frac12\right)^2+\frac34}\right] \end{align*} then \begin{align*} \sum_{k=1}^{99}\frac{k}{1+k^2+k^4}&=\frac12\sum_{k=1}^{99}\left[\frac{1}{\left(k-\frac12\right)^2+\frac34}-\frac{1}{\left(k+\frac12\right)^2+\frac34}\right] \end{align*} observe that the last sum is telescopic, so $$\sum_{k=1}^{99}\frac{k}{1+k^2+k^4}=\frac12\left[\frac{1}{\left(1-\frac12\right)^2+\frac34}-\frac{1}{\left(99+\frac12\right)^2+\frac34}\right]=\frac{4950}{9901}\approx 0.4999495$$
Hence the answer is $\color{blue}{D)}$.
You can consider the fact that:
$$\frac{x}{1+x^2+x^4}=\frac{x}{(1+2x^2+x^4)-x^2}=\frac{x}{(x^2+1)^2-x^2}=\frac{x}{(x^2+x+1)(x^2+1-x)}$$
Then, considering the last one, you have:
$$\frac{x}{(x^2+x+1)(x^2+1-x)}=\frac{1/2}{x^2-x+1}-\frac{1/2}{x^2+x+1}$$
From where you can finish using a telescopic sum.
Answer is D.
We can write it as $\frac{r}{1+r^2+r^4}$. So now we can write it as $\frac{1}{2}(\frac{1}{1-r+r^2}-\frac{1}{1+r+r^2})$ And after putting the values from 1 to 99 we will got 0.4999449
