Sequences & Series
Sequences and Series
nta_pyq_2025_jan
Grade 11
Question:
If $\displaystyle\sum_{r=1}^{n}T_{r}=\dfrac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$, then $\displaystyle\lim_{n\to\infty}\sum_{r=1}^{n}\dfrac{1}{T_{r}}$ is equal to:
$\dfrac{2}{3}$
0
1
$\dfrac{1}{3}$
Step-by-Step Solution
Key Concept: $T_{n}=S_{n}-S_{n-1}$. The product structure makes $T_{n}=\dfrac{(2n-1)(2n+1)(2n+3)}{8}$. Then partial fractions split $\dfrac{1}{(2r-1)(2r+1)(2r+3)}$ into a telescoping difference.
$T_{n}=S_{n}-S_{n-1}=\dfrac{(2n-1)(2n+1)(2n+3)}{64}\bigl[(2n+5)-(2n-3)\bigr]=\dfrac{(2n-1)(2n+1)(2n+3)}{8}.$
Use $\dfrac{1}{(2r-1)(2r+1)(2r+3)}=\dfrac{1}{4}\!\left[\dfrac{1}{(2r-1)(2r+1)}-\dfrac{1}{(2r+1)(2r+3)}\right].$
So $\dfrac{1}{T_{r}}=\dfrac{8}{(2r-1)(2r+1)(2r+3)}=2\!\left[\dfrac{1}{(2r-1)(2r+1)}-\dfrac{1}{(2r+1)(2r+3)}\right].$
Telescoping:
$$\sum_{r=1}^{n}\frac{1}{T_{r}}=2\!\left[\frac{1}{1\cdot 3}-\frac{1}{(2n+1)(2n+3)}\right]\xrightarrow{n\to\infty}\frac{2}{3}.$$
Correct Answer: 1