If the sum of the first 20 terms of the series 4.1/4+3.1^2+1^4 + 4.2/4+3.2^2+2^4 + 4.3/4+3.3^2+3^4 + 4.4/4+3.4^2+4^4 + .... is m/n,

Question

If the sum of the first 20 terms of the series \(\frac{4.1}{4+3.1^{2}+1^{4}}+\frac{4.2}{4+3.2^{2}+2^{4}}+\frac{4.3}{4+3.3^{2}+3^{4}}+\frac{4.4}{4+3.4^{2}+4^{4}}+\ldots.\) is \(\frac{m}{n},\) where m and n are coprime, then m + n is equal to :-

(1) 423

(2) 420

(3) 421

(4) 422  

Answer 1

Correct option is: (3) 421 

\(S_n=\sum\limits_{r=1}^n \frac{4 r}{4+3 r^2+r^4} \)

\(=2 \sum\limits_{r=1}^n \frac{2 r}{\left.(r^2+2\right)^2-r^2}=2 \sum\limits_{r=1}^n \frac{\left(r^2+2+r\right)-\left(r^2+2-r\right)}{\left(r^2+2+r\right)\left(r^2+2-r\right)} \)

\(=2 \sum\limits_{r=1}^n\left(\frac{1}{r^2+2-r}-\frac{1}{r^2+2+r}\right) \)

\(S_{20}=2\left[\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+\ldots\right] \)

\(=2\left(\frac{1}{2}-\frac{1}{20^2+2+20}\right) \)

\(=2\left(\frac{1}{2}-\frac{1}{422}\right) \)

\(=2\left(\frac{422-2}{422 \times 2}\right)=\frac{420}{422}=\frac{210}{211}=\frac{m}{n} \)

\(m+n=421\)