JEE Main 2023 (Online) 30th January Morning ShiftQuestion 1
If $a_n=\frac{-2}{4n^2-16n+15}$, then $a_1+a_2+\ldots+a_{25}$ is equal to:
Sequence and Series · Telescoping ($V_n$ Method)
Solution
Formulas used. (1) $4n^2-16n+15=(2n-5)(2n-3)$(2) $\frac{-2}{(2n-5)(2n-3)}=\frac1{2n-3}-\frac1{2n-5}$Step 1 - Split. Check: $\dfrac1{2n-3}-\dfrac1{2n-5}=\dfrac{(2n-5)-(2n-3)}{(2n-3)(2n-5)}=\dfrac{-2}{(2n-3)(2n-5)}$. Step 2 - Telescope. Put $W_n=\dfrac1{2n-5}$; then $\dfrac1{2n-3}=W_{n+1}$ and $a_n=W_{n+1}-W_n$: $$\sum_{n=1}^{25}a_n=W_{26}-W_1=\frac1{47}-\frac1{-3}=\frac1{47}+\frac13 .$$ Step 3. $$\frac{3+47}{141}=\frac{50}{141}.$$ Answer: (C)