If $\sum_{r=1}^nT_r=\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$, then $\lim_{n o\infty}\sum_{r=1}^n\left(\frac1{T_r} ight)$ is equal to :

Question

If $\sum_{r=1}^nT_r=\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$, then $\lim_{n	o\infty}\sum_{r=1}^n\left(\frac1{T_r}ight)$ is equal to :

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $T_n=S_n-S_{n-1}=\frac{(2n-1)(2n+1)(2n+3)}8$, so $\frac1{T_n}=2\left[\frac1{(2n-1)(2n+1)}-\frac1{(2n+1)(2n+3)}\right]$ and the sum telescopes to $2\cdot\frac13=\frac23$. (Medium confidence: strictly, $T_1=S_1=\frac{105}{64}$ differs from the general-term formula, and a literal sum would give $\frac{26}{35}$, which is not an option; the official key follows the telescoping result.) Answer (2).

Q. If $\sum_{r=1}^nT_r=\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$, then $\lim_{n\to\infty}\sum_{r=1}^n\left(\frac1{T_r}\right)$ is equal to :

Explanation

Reviewed by Dr. Abubacker

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