Question

Let $S=2+\frac{6}{7}+\frac{12}{7 ^2}+\frac{20}{7 ^3}+\frac{30}{7 ^4}+…$ Then $4S$ is equal to

• A. $\bigg(\frac{7}{3}\bigg)^2$
• B. $\bigg(\frac{7 ^3}{3 ^2}\bigg)$
• C. $\bigg(\frac{7}{3}\bigg)^3$
• D. $\bigg(\frac{7 ^2}{3^ 3}\bigg)$

Answer 1

Answer: (C)

$\bigg(\frac{7}{3}\bigg)^3$

Answer 2

$S=2+\frac{6}{7}+\frac{12}{7 ^2}+\frac{20}{7 ^3}+\frac{30}{7 ^4}+…$ …(i)

$\frac{1}{7}S=\frac{2}{7}+\frac{6}{7^ 2}+\frac{12}{7 ^3}+\frac{20}{7 ^4}+$..…(ii)

$(i) – (ii)$

$\frac{6}{7}S=2+\frac{4}{7}+\frac{6}{7 ^2}+\frac{8}{7 ^3}+.$.…(iii)

$\frac{6}{7 ^2}S=\frac{2}{7}+\frac{4}{7^ 2}+\frac{6}{7 ^3}+$……(iv)

$(iii) – (iv)$

$(\frac{6}{7})^2S=2+\frac{2}{7}+\frac{2}{7^2}+\frac{2}{7^3}+..$

=$2\bigg[\frac{1}{\bigg(1−\frac{1}{7}\bigg)}]=2\bigg(\frac{7}{6}\bigg)$

$∴ \;4S=8\bigg(\frac{7}{6}\bigg)^3=\bigg(\frac{7}{3}\bigg)^3$