Question

\( 1 + \frac{4}{15} + \frac{4 \cdot 10}{15 \cdot 30} + \frac{4 \cdot 10 \cdot 16}{15 \cdot 30 \cdot 45} + \cdots \infty \)

• A. \( \left( \frac{3}{5} \right)^{2/3} \)
• B. \( \left( \frac{5}{3} \right)^{2/3} \)
• C. \( \left( \frac{3}{5} \right)^{3/2} \)
• D. \( \left( \frac{5}{3} \right)^{3/2} \)

Hint:

For infinite series with patterns, identify the general term by examining the numerators and denominators separately, then use known series expansions like the logarithm series to simplify.

Answer 1

The Correct Option is B

Step 1: Identify the Pattern
The given series can be written as:
\[ S = 1 + \sum_{n=1}^\infty \frac{\prod_{k=1}^n (6k-2)}{\prod_{k=1}^n (15k)} \] Step 2: Rewrite Using Pochhammer Symbols
Express the products using Pochhammer's symbol:
\[ S = \sum_{n=0}^\infty \frac{(\frac{2}{3})_n (\frac{1}{3})_n}{n!} \left(\frac{4}{5}\right)^n \] Step 3: Recognize as Hypergeometric Series
This is a hypergeometric series:
\[ S = {}_2F_1\left(\frac{2}{3}, \frac{1}{3}; 1; \frac{4}{5}\right) \] Step 4: Apply Hypergeometric Identity
Using the identity:
\[ {}_2F_1(a,b;a+b+\frac{1}{2};z) = {}_2F_1(2a,2b;a+b+\frac{1}{2};\frac{1-\sqrt{1-z}}{2}) \]
and evaluating at \( z = \frac{4}{5} \), we get:
\[ S = \left( \frac{5}{3} \right)^{2/3} \]