Question

The sum $\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+...+\frac{99}{1+{99}^2+{99}^4}$ lies between

• A. $0.46$ and $0.47$
• B. $0.52$ and $1.0$
• C. $0.48$ and $0.49$
• D. $0.49$ and $0.50$

Answer 1

The correct option is A $0.46$ and $0.47$

We have,

$\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1^3+2^2+3^4}+.........+\frac{99}{1+{99}^2+{99}^4}={\sum }_{n=1}^{99}\frac{x}{1+x^2+x^4}={\sum }_{n=1}^{99}\frac{x}{{\left(x^2+1\right)}^2-x^2}={\sum }_{n=1}^{99}\frac{\left(x^2+x+1\right)-\left(x^2-x+1\right)}{2\left(x^2+x+1\right)\left(x^2-x+1\right)}=\frac{1}{2}{\sum }_{n=1}^{99}\left(\frac{1}{x^2-x+1}-\frac{1}{x^2+x+1}\right)=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{7}+\frac{1}{7}+.........+\frac{1}{{99}^2+99+1}\right)=\frac{1}{2}\left(1-\frac{1}{9901}\right)=\frac{4950}{9901}\approx \frac{1}{2}$

$Hence,optionDiscorrectanswer.$