Question

If the value of \(\left( 1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + ....... \text{upto } \infty \right)^{\log_{0.25} \left( \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + ....... \text{upto } \infty \right)}\) is \(l\), then \(l^2\) is equal to __________.

Hint:

For an infinite AGP, write the sum $S$, multiply it by the common ratio $r$, and subtract $S - rS$. This method is more reliable than memorizing formulas.

Answer 1

Correct Answer: 3

Step 1: Understanding the Concept:
The problem consists of an infinite Arithmetico-Geometric Progression (AGP) and an infinite Geometric Progression (GP) inside a logarithm. We solve these series first to simplify the expression.
Step 2: Key Formula or Approach:
1. Sum of infinite GP: \(S_{\infty} = \frac{a}{1-r}\).
2. Sum of infinite AGP: \(S = a + \frac{(a+d)r}{1-r} + ....... = \frac{a}{1-r} + \frac{dr}{(1-r)^2}\).
Step 3: Detailed Explanation:
1. Evaluate the base AGP (\(S_1\)):
\(S_1 = 1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + .......\)
Let \(S' = \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + .......\)
This is an AGP with \(a=2, d=4, r=1/3\).
\[ S' = \frac{2/3}{1-1/3} + \frac{4(1/3)}{(1-1/3)^2} = \frac{2/3}{2/3} + \frac{4/3}{4/9} = 1 + 3 = 4 \]
So, \(S_1 = 1 + S' = 1 + 2 = 3\). (Wait, re-checking: $S' = \frac{2}{3} + \frac{6}{9} + ....... = \frac{2}{3} + \frac{2}{3} + .......$, pattern is from 2nd term).
Let's use subtraction: \(S_1 = 1 + \frac{2}{3} + \frac{6}{9} + \frac{10}{27} + .......\)
\(\frac{1}{3}S_1 = \frac{1}{3} + \frac{2}{9} + \frac{6}{27} + .......\)
\(S_1(1-1/3) = 1 + \frac{1}{3} + \frac{4}{9} + \frac{4}{27} + .......\)
\(\frac{2}{3}S_1 = \frac{4}{3} + \frac{4/9}{1-1/3} = \frac{4}{3} + \frac{4/9}{2/3} = \frac{4}{3} + \frac{2}{3} = 2\)
\(S_1 = 2 \times \frac{3}{2} = 3\).

2. Evaluate the inner GP (\(S_2\)):
\(S_2 = \frac{1}{3} + \frac{1}{3^2} + ....... = \frac{1/3}{1-1/3} = \frac{1/3}{2/3} = \frac{1}{2}\).

3. Evaluate the exponent:
Exponent \(= \log_{0.25} (1/2) = \log_{(1/2)^2} (1/2) = \frac{1}{2}\).

4. Find \(l\) and \(l^2\):
\(l = 3^{1/2} = \sqrt{3}\).
\(l^2 = 3\).
Step 4: Final Answer:
The value of \(l^2\) is 3.