Find the sum :

We can split the above expression into 2 parts. We will split 2n terms into 2 parts also which will leave it as n terms and another n terms.

$=\left(\frac{3}{5}+\frac{3}{5^{3}}+\ldots\right.$ to n terms $)+\left(\frac{4}{5}+\frac{4}{5^{2}}+\ldots\right.$ to n terms $)$

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,

$a=\frac{3}{5}, \frac{4}{5}$

r = (ratio between the n term and n-1 term) $\frac{3}{5^{3}} \div \frac{3}{5}, \frac{4}{5^{2}} \div \frac{4}{5}=\frac{1}{5^{2}}, \frac{1}{5}$

n terms

$\therefore \mathrm{S}_{\mathrm{n}}=\frac{3}{5} \times \frac{1-\frac{1}{5^{2}}^{\mathrm{n}}}{1-\frac{1}{5^{2}}}+\frac{4}{5} \times \frac{1-\frac{1}{5}^{\mathrm{n}}}{1-\frac{1}{5}}$

$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{3}{5} \times \frac{1-\frac{1}{5^{2}}}{\frac{24}{5^{2}}}+\frac{4}{5} \times \frac{1-\frac{1}{5}^{\mathrm{n}}}{\frac{4}{5}}$

$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{5\left(1-\frac{1}{5^{2}}^{\mathrm{n}}\right)}{8}+\left(1-\frac{1}{5}^{\mathrm{n}}\right)$

$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\left(5-\frac{5}{5^{2 n}}\right)}{8}+\left(1-\frac{1}{5}^{\mathrm{n}}\right)$

$\therefore \mathrm{S}_{\mathrm{n}}=\frac{\left(5-\frac{1}{5^{2 n-1}}\right)}{8}+\left(1-\frac{1^{n}}{5}\right)$