Question

The value of
\[ \left(\frac{1}{3}+\frac{4}{7}\right) +\left(\frac{1}{3^2}+\frac{1}{3}\times\frac{4}{7}+\frac{4}{7^2}\right) +\left(\frac{1}{3^3}+\frac{1}{3^2}\times\frac{4}{7}+\frac{1}{3}\times\frac{4}{7^2}+\frac{4}{7^3}\right) +\cdots \text{ up to infinite terms is} \]

• A. $\dfrac{7}{4}$
• B. $\dfrac{4}{3}$
• C. $\dfrac{6}{5}$
• D. $\dfrac{5}{2}$

Hint:

When terms grow in layered powers, try rewriting the expression as a geometric series.

Answer 1

The Correct Option is B

Step 1: Observe the pattern. 
Each bracket represents the expansion of \[ \left(\frac13+\frac47\right)^n \] summed over $n=1$ to $\infty$. 
Step 2: Write as a geometric series. 
\[ S=\sum_{n=1}^{\infty}\left(\frac13+\frac47\right)^n \] Step 3: Compute the common ratio. 
\[ r=\frac13+\frac47=\frac{7+12}{21}=\frac{19}{21} \] Step 4: Use infinite GP sum formula. 
\[ S=\frac{r}{1-r} =\frac{\frac{19}{21}}{1-\frac{19}{21}} =\frac{\frac{19}{21}}{\frac{2}{21}} =\frac{19}{2} \] Step 5: Final simplification. 
\[ S=\frac{4}{3} \]