The sum to infinity of the series \(1+\frac{2}{3}+\frac{6}{3^2}+\frac{

1.	The sum to infinity of the series \(1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+......\)is(a) 6 (b) 2 (c) 3 (d) 4
Answer» (b) 2 S = \(1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+......\) upto ∞ \(\frac{1}{3}S\) = \(\frac{1}{3}+\frac{2}{3^2}+\frac{6}{3^3}+\frac{10}{3^4}+......\)upto ∞ ⇒ S - \(\frac{1}{3}S\) = \(1+\frac{1}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+......\)upto ∞ = \(\frac{4}{3}\) + 4 \(\bigg[\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+.......\infty\bigg]\) = \(\frac{4}{3}\) + 4 \(\bigg[\frac{\frac{1}{3^3}}{1-\frac{1}{3}}\bigg]\)\(\big(\because\,S_\infty=\frac{a}{1-r}\big)\) = \(\frac{4}{3}\) + 4 \(\bigg[\frac{\frac{1}{9}}{\frac{2}{3}}\bigg]\) = \(\frac{4}{3}\) + \(\frac{4}{6}\) = \(\frac{12}{6}\) = 2.

The sum to infinity of the series \(1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+......\)is(a) 6 (b) 2 (c) 3 (d) 4

Answer»

(b) 2

S = \(1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+......\) upto ∞

\(\frac{1}{3}S\) = \(\frac{1}{3}+\frac{2}{3^2}+\frac{6}{3^3}+\frac{10}{3^4}+......\)upto ∞

⇒ S - \(\frac{1}{3}S\) = \(1+\frac{1}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+......\)upto ∞

= \(\frac{4}{3}\) + 4 \(\bigg[\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+.......\infty\bigg]\)

= \(\frac{4}{3}\) + 4 \(\bigg[\frac{\frac{1}{3^3}}{1-\frac{1}{3}}\bigg]\)\(\big(\because\,S_\infty=\frac{a}{1-r}\big)\)

= \(\frac{4}{3}\) + 4 \(\bigg[\frac{\frac{1}{9}}{\frac{2}{3}}\bigg]\) = \(\frac{4}{3}\) + \(\frac{4}{6}\) = \(\frac{12}{6}\) = 2.