The value of
\[
\left(\frac{1}{3}+\frac{4}{7}\right)
+\left(\frac{1}{3^2}+\frac{1}{3}\times\frac{4}{7}+\frac{4}{7^2}\right)
+\left(\frac{1}{3^3}+\frac{1}{3^2}\times\frac{4}{7}+\frac{1}{3}\times\frac{4}{7^2}+\frac{4}{7^3}\right)
+\cdots \text{ up to infinite terms is}
\]
\[ \left(\frac{1}{3}+\frac{4}{7}\right) +\left(\frac{1}{3^2}+\frac{1}{3}\times\frac{4}{7}+\frac{4}{7^2}\right) +\left(\frac{1}{3^3}+\frac{1}{3^2}\times\frac{4}{7}+\frac{1}{3}\times\frac{4}{7^2}+\frac{4}{7^3}\right) +\cdots \text{ up to infinite terms is} \]
Show Hint
- $\dfrac{7}{4}$
- $\dfrac{4}{3}$
- $\dfrac{6}{5}$
- $\dfrac{5}{2}$
The Correct Option is B
Solution and Explanation
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.
\[ S=\frac{4}{3} \]
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