Question

The sum of infinite series $\frac{5}{\angle 3}+\frac{19}{\angle 5}+\frac{41}{\angle 7}+\frac{71}{\angle 9}+\dots \dots$ is

• A. $\frac{3}{2}$
• B. $1$
• C. $\frac{9}{2}$
• D. $2$

Answer 1

The correct option is B $1$

$T_r=\frac{4r^2+2r-1}{\angle (2r+1)}=\frac{2r(2r+1)-1}{\angle 2r+1}$
$=\frac{1}{\angle 2r-1}-\frac{1}{\angle 2r+1}$
$T_1+T_2+\dots \dots T_n=(\frac{1}{\angle 1}-\frac{1}{\angle 3})+(\frac{1}{\angle 3}-\frac{1}{\angle 5})+\dots \dots +\frac{1}{\angle 2n-1}-\frac{1}{\angle 2n+1}$
$=1-\frac{1}{\angle 2n+1}asn\to \infty S_{\infty }\to 1$
Hence sum = 1