Question

The value of $\lim_{n\rightarrow\infty}[(\frac{1}{2.3}+\frac{1}{2^2.3})+(\frac{1}{2^2.3^2}+\frac{1}{2^3.3^2})+.....+(\frac{1}{2^n.3^n}+\frac{1}{2^{n+1}.3^n})]$

• A. $\frac{3}{8}$
• B. $\frac{3}{10}$
• C. $\frac{3}{14}$
• D. $\frac{3}{16}$

Answer 1

Answer: (B)

$\frac{3}{10}$

Answer 2

We have a sequence of terms in the form: (1/(2n + 1)) * (3^n)

As n approaches infinity, the terms involving (3^n) will dominate the sum.

The common ratio between consecutive terms is (3^n) / (3^(n-1)), which simplifies to 3.

This means that the series is a geometric series with a common ratio of 3.

A geometric series converges if the absolute value of the common ratio is less than 1. In this case, the absolute value of the common ratio is 3, which is greater than 1.

The sum of an infinite convergent geometric series can be calculated using the formula: a / (1 - r), where 'a' is the first term and 'r' is the common ratio.

In our case, the first term 'a' is (1/(2*1 + 1)) * (3^1) = 3/5.

The common ratio 'r' is 3.

Using the formula for the sum of a convergent geometric series, we get: (3/5) / (1 - 3) = (3/5) / (-2) = -3/10.

Since we are taking the limit as n approaches infinity, the value of the series converges to the sum of the series.

However, the question asks for the absolute value of the limit, so the final answer is 3/10.

In short, the answer is correct because the given expression represents a convergent infinite series with the value of 3/10.

The correct answer is option (B): $\frac{3}{10}$