Question:
Find the sum of the GP :
$1-\frac{1}{3}+\frac{1}{3^{2}}-\frac{1}{3^{3}}+$ … to n terms
Solution:
Sum of a G.P. series is represented by the formula $S_{n}=a \frac{1-r^{n}}{1-r}$
when |r|<1. ‘Sn’ represents the sum of the G.P. series upto nth terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.
Here
a = 1
r = (ratio between the n term and n-1 term) $-\frac{1}{3} \div 1=-\frac{1}{3}$
n terms
$\therefore S_{\mathrm{n}}=1 \times \frac{1-\frac{-1}{3}^{\mathrm{n}}}{1-\frac{1}{3}}$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{1-\frac{1}{3}^{\mathrm{n}}}{\frac{2}{3}}$
$\therefore \mathrm{S}_{\mathrm{n}}=\frac{3-\frac{1}{3}^{\mathrm{n}-1}}{2}$