JEE Main 2022 · 25 Jul · Shift 2 · Mathematics Q5

MathematicsSingle correct+4 / -1Question 5 of 90

The sum $\displaystyle\sum_{n=1}^{21}\frac3{(4n-1)(4n+3)}$ is equal to

\dfrac7{87}

\dfrac7{29}

\dfrac{14}{87}

\dfrac{21}{29}

Answer and worked solution

Official answer: B

Telescoping. $\dfrac3{(4n-1)(4n+3)}=\dfrac34\left(\dfrac1{4n-1}-\dfrac1{4n+3}\right)$ ; sum $\frac34\left(\frac13-\frac1{87}\right)=\frac7{29}$ .

(B)

Topics:Special series sums

JEE Main 2022 · 25 Jul · Shift 2, Mathematics question 5