Progression (Sequence and Series)HardBloom L3
Question
The sum $\frac{7}{2 \times 3}\left( \frac{1}{3} \right) + \frac{9}{3 \times 4}\left( \frac{1}{3} \right)^{2} + \frac{11}{4 \times 5}\left( \frac{1}{3} \right)^{3} + \frac{13}{5 \times 6}\left( \frac{1}{3} \right)^{4} + \ldots$. upto 10 terms is equal to
Options
A.$1 - \frac{1}{12 \times 3^{10}}$
B.$\frac{1}{2} - \frac{1}{12 \times 3^{10}}$
C.$\frac{1}{3} - \frac{1}{12 \times 3^{10}}$
D.$\frac{1}{2} - \frac{1}{11 \times 3^{10}}$
Solution
Sol. [Key Concept: Arithmetic–geometric / telescoping series in powers of $\frac{1}{13}$.]
$$\sum_{r = 1}^{10}\mspace{2mu}\frac{2r + 5}{(r + 1)(r + 2)}\frac{1}{3^{r}} = \sum_{r = 1}^{10}\mspace{2mu}\frac{3(r + 2) - (r + 1)}{(r + 1)(r + 2)}\frac{1}{3^{r}}$$
$$\begin{matrix} & \ = \sum_{r = 1}^{10}\mspace{2mu}\mspace{2mu}\left( \frac{1}{(r + 1)3^{r - 1}} - \frac{1}{(r + 2)3^{r}} \right) \\ & \ = \frac{1}{2} - \frac{1}{12 \cdot 3^{10}} \end{matrix}$$
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