JEE Main 2022 (Online) 25th July Evening ShiftQuestion 1
The sum $\sum_{n=1}^{21}\frac{3}{(4n-1)(4n+3)}$ is equal to
Sequence and Series · Telescoping ($V_n$ Method)
Solution
Formulas used. (1) $\frac{1}{(4n-1)(4n+3)}=\frac14\left[\frac1{4n-1}-\frac1{4n+3}\right]$Step 1 - Split. $$\frac3{(4n-1)(4n+3)}=\frac34\left[\frac1{4n-1}-\frac1{4n+3}\right].$$ Step 2 - Telescope ($n=1$ to $21$): $\dfrac13-\dfrac17+\dfrac17-\dfrac1{11}+\cdots-\dfrac1{87}$: $$\frac34\left(\frac13-\frac1{87}\right)=\frac34\cdot\frac{28}{87}=\frac{21}{87}=\frac7{29}.$$ Answer: (B)