MCQ
Let $S=2+\frac{6}{7}+\frac{12}{7^{2}}+\frac{20}{7^{3}}+\frac{30}{7^{4}}+\ldots . .$ then $4 S$ is equal to
- A$\left(\frac{7}{3}\right)^{2}$
- B$\frac{7^{3}}{3^{2}}$
- ✓$\left(\frac{7}{3}\right)^{3}$
- D$\frac{7^{2}}{3^{3}}$
$S =2+\frac{6}{7}+\frac{12}{7^{2}}+\frac{20}{7^{3}}+\frac{30}{7^{4}}+\ldots \ldots \ldots$
$\frac{S}{7}=\frac{2}{7}+\frac{6}{7^{2}}+\frac{12}{7^{3}}+\frac{20}{7^{4}}+\ldots \ldots \ldots$
$\Rightarrow \frac{6 S }{7}=2+\frac{4}{7}+\frac{6}{7^{2}}+\frac{8}{7^{3}}+\frac{10}{7^{4}}+\ldots \ldots$
$\Rightarrow \frac{6 S }{7^{2}}=\quad \frac{2}{7}+\frac{4}{7^{2}}+\frac{6}{7^{3}}+\frac{8}{7^{4}}+\ldots \ldots \ldots$
$\Rightarrow \frac{6 S }{7}\left(1-\frac{1}{7}\right)=2+\frac{2}{7}+\frac{2}{7^{2}}+\frac{2}{7^{3}}+\ldots \ldots \ldots$
$\Rightarrow \frac{6^{2} S }{7^{2}}=\frac{2}{1-\frac{1}{7}}=\frac{2}{6} \times 7$
$\Rightarrow S= \frac{2 \times 7^{3}}{6^{3}} \Rightarrow 4 S =\frac{7^{3}}{3^{3}}=\left(\frac{7}{3}\right)^{3}$
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