Question

The value of \(1+\bigg(1+\frac{1}{3}\bigg)\frac{1}{4}+\bigg(1+\frac{1}{3}+\frac{1}{9}\bigg)\frac{1}{16}+\bigg(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\bigg)\frac{1}{64}+....\),is

• A. \(\frac{15}{13}\)
• B. \(\frac{27}{12}\)
• C. \(\frac{15}{11}\)
• D. \(\frac{15}{8}\)

Answer 1

The Correct Option is C

This problem involves evaluating a summation series: 

\[ 1 + \left(1 + \frac{1}{3}\right)\frac{1}{4} + \left(1 + \frac{1}{3} + \frac{1}{9}\right)\frac{1}{16} + \left(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}\right)\frac{1}{64} + \cdots \]

The expression inside the parentheses is a geometric series with first term \( a = 1 \) and common ratio \( r = \frac{1}{3} \), up to \( n \) terms. The sum of a finite geometric series is:

\[ S_n = \frac{a(1 - r^n)}{1 - r} \Rightarrow \frac{1 - \left(\frac{1}{3}\right)^n}{1 - \frac{1}{3}} = \frac{1 - \left(\frac{1}{3}\right)^n}{\frac{2}{3}} = \frac{3}{2} \left(1 - \left(\frac{1}{3}\right)^n\right) \]

Each of these is multiplied by \( \frac{1}{4^n} \), so the full series becomes:

\[ 1 + \sum_{n=1}^{\infty} \left( \frac{3}{2} \left(1 - \left(\frac{1}{3}\right)^n \right) \cdot \frac{1}{4^n} \right) \]

We split the sum into two parts:

\[ 1 + \frac{3}{2} \left( \sum_{n=1}^{\infty} \frac{1}{4^n} - \sum_{n=1}^{\infty} \frac{1}{(4 \cdot 3)^n} \right) \Rightarrow 1 + \frac{3}{2} \left( \sum_{n=1}^{\infty} \frac{1}{4^n} - \sum_{n=1}^{\infty} \frac{1}{12^n} \right) \]

Each of these is a geometric series:

\[ \sum_{n=1}^{\infty} \frac{1}{4^n} = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{1}{3}, \quad \sum_{n=1}^{\infty} \frac{1}{12^n} = \frac{\frac{1}{12}}{1 - \frac{1}{12}} = \frac{1}{11} \]

Substitute back into the expression:

\[ 1 + \frac{3}{2} \left( \frac{1}{3} - \frac{1}{11} \right) = 1 + \frac{3}{2} \cdot \frac{8}{33} = 1 + \frac{24}{66} = 1 + \frac{12}{33} = \frac{33 + 12}{33} = \frac{45}{33} = \frac{15}{11} \]

Correction: The correct final simplified result is actually:

\[ \frac{33 + 24}{33} = \frac{57}{33} = \frac{19}{11} \]

But your earlier math concluded:

\[ 1 + \frac{24}{66} = \frac{22 + 8}{22} = \frac{30}{22} = \frac{15}{11} \]

Which is correct if that is what the simplification gives, but since we are dealing with:

\[ 1 + \frac{3}{2} \left( \frac{1}{3} - \frac{1}{11} \right) = 1 + \frac{3}{2} \cdot \frac{8}{33} = 1 + \frac{24}{66} = \frac{90}{66} = \frac{15}{11} \]

✅ Final Answer: \(\boxed{\frac{15}{11}}\)