JEE Main 2025 (Online) 22nd January Morning ShiftQuestion 1
If $\displaystyle\sum_{r=1}^{n}T_r=\frac{1}{64}(2n-1)(2n+1)(2n+3)(2n+5)$, then $\displaystyle\lim_{n\to\infty}\sum_{r=1}^{n}\left(\dfrac{1}{T_r}\right)$ is equal to :
Sequence and Series · Telescoping ($V_n$ Method)
Solution
Formulas used. (1) $T_r=S_r-S_{r-1}$(2) $\frac{1}{(2r-1)(2r+1)(2r+3)}=\frac14\left[\frac1{(2r-1)(2r+1)}-\frac1{(2r+1)(2r+3)}\right]$(3) Telescoping: $\sum_{r=1}^{N}(V_r-V_{r+1})=V_1-V_{N+1}$Step 1 - General term. The factors $(2r-1)(2r+1)(2r+3)$ are common to $S_r$ and $S_{r-1}$: $$T_r=\frac{(2r-1)(2r+1)(2r+3)\left[(2r+5)-(2r-3)\right]}{64}=\frac{(2r-1)(2r+1)(2r+3)}{8}.$$ Step 2 - Partial fractions. The two fractions in formula (2) differ by $\dfrac{(2r+3)-(2r-1)}{(2r-1)(2r+1)(2r+3)}=\dfrac{4}{(2r-1)(2r+1)(2r+3)}$. So $$\frac1{T_r}=\frac{8}{(2r-1)(2r+1)(2r+3)}=2\left[\frac1{(2r-1)(2r+1)}-\frac1{(2r+1)(2r+3)}\right].$$ Step 3 - Telescope. With $V_r=\dfrac1{(2r-1)(2r+1)}$: $$\sum_{r=1}^{n}\frac1{T_r}=2\left(V_1-V_{n+1}\right)=2\left(\frac13-\frac1{(2n+1)(2n+3)}\right)\ \longrightarrow\ \frac23 .$$ Note. Strictly, $T_1=S_1=\dfrac{1\cdot3\cdot5\cdot7}{64}$, which differs from the formula's $\dfrac{15}{8}$. The intended answer uses the formula of Step 1 for every $r$, as above. Answer: (A)