Let S=2+6 7 +12 7^ 2 +20 7^ 3 +30 7^ 4 + . . then 4 S is equal to - Vidyadip

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
$\frac{7^{2}}{3^{3}}$

✓

Answer

Correct option: C.

$\left(\frac{7}{3}\right)^{3}$

c
$S=2+\frac{6}{7}+\frac{12}{7^{2}}+\frac{20}{7^{3}}+\frac{30}{7^{4}}+\ldots \ldots$ Considering infinite sequence,

$S =2+\frac{6}{7}+\frac{12}{7^{2}}+\frac{20}{7^{3}}+\frac{30}{7^{4}}+\ldots \ldots \ldots$

$\frac{S}{7}=\frac{2}{7}+\frac{6}{7^{2}}+\frac{12}{7^{3}}+\frac{20}{7^{4}}+\ldots \ldots \ldots$

$\Rightarrow \frac{6 S }{7}=2+\frac{4}{7}+\frac{6}{7^{2}}+\frac{8}{7^{3}}+\frac{10}{7^{4}}+\ldots \ldots$

$\Rightarrow \frac{6 S }{7^{2}}=\quad \frac{2}{7}+\frac{4}{7^{2}}+\frac{6}{7^{3}}+\frac{8}{7^{4}}+\ldots \ldots \ldots$

$\Rightarrow \frac{6 S }{7}\left(1-\frac{1}{7}\right)=2+\frac{2}{7}+\frac{2}{7^{2}}+\frac{2}{7^{3}}+\ldots \ldots \ldots$

$\Rightarrow \frac{6^{2} S }{7^{2}}=\frac{2}{1-\frac{1}{7}}=\frac{2}{6} \times 7$

$\Rightarrow S= \frac{2 \times 7^{3}}{6^{3}} \Rightarrow 4 S =\frac{7^{3}}{3^{3}}=\left(\frac{7}{3}\right)^{3}$

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