WBJEE · Maths · Sequences and Series
The value of \(\lim _{n \rightarrow \infty}\left[\left(\frac{1}{2 \cdot 3}+\frac{1}{2^2 \cdot 3}\right)+\left(\frac{1}{2^2 \cdot 3^2}+\frac{1}{2^3 \cdot 3^2}\right)+\ldots+\left(\frac{1}{2^n \cdot 3^n}+\frac{1}{2^{n+1} \cdot 3^n}\right)\right]\) is
- A \(\frac{3}{8}\)
- B \(\frac{3}{10}\)
- C \(\frac{3}{14}\)
- D \(\frac{3}{16}\)
Answer & Solution
Correct Answer
(B) \(\frac{3}{10}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} \text{Hint : } & \lim _{n \rightarrow \infty} \sum_{x=1}^n\left(\frac{1}{2^x \times 3^x}+\frac{1}{2^{x+1} \times 3^x}\right) \\ & =\lim _{n \rightarrow \infty} \sum_{x=1}^n \frac{1}{6^x}\left(1+\frac{1}{2}\right) \\ & =\frac{3}{2} \times \underbrace{\lim _{n…
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