Evaluate :

We can write this as $\left(2+3^{1}\right)+\left(2+3^{2}\right)+\left(2+3^{3}\right)+\ldots$ to 10 terms

$=(2+2+2+\ldots$ to 10 terms $)+\left(3+3^{2}+3^{3}+\ldots\right.$ to 10 terms $)$

$=2 \times 10+\left(3+3^{2}+3^{3}+\ldots\right.$ to 10 terms $)$

$=20+\left(3+3^{2}+3^{3}+\ldots\right.$ to 10 terms $)$

Sum of a G.P. series is represented by the formula $\mathrm{S}_{\mathrm{n}}=\mathrm{a} \frac{\mathrm{r}^{\mathrm{n}}-1}{\mathrm{r}-1}$ 

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 = 3

r = (ratio between the n term and n-1 term) 3

n = 10 terms

$\mathrm{S}_{\mathrm{n}}=3 \times \frac{3^{10}-1}{3-1}$

$\Rightarrow \mathrm{S}_{\mathrm{n}}=3 \times \frac{59049-1}{2}$

$\Rightarrow \mathrm{S}_{\mathrm{n}}=3 \times \frac{59048}{2}$

$\Rightarrow \mathrm{S}_{\mathrm{n}}=88572$

Thus, sum of the given expression is

$=20+\left(3+3^{2}+3^{3}+\ldots\right.$ to 10 terms $)$

$=20+88572$

$=88592$

(ii) The given expression can be written as,

$\left(2^{1}+3^{1-1}\right)+\left(2^{2}+3^{2-1}\right)+\ldots$ to $n$ terms

$=\left(2+3^{0}\right)+\left(2^{2}+3^{1}\right)+\ldots$ to $n$ terms

$=(2+1)+\left(2^{2}+3\right)+\ldots$ to $n$ terms

$=\left(2+2^{2}+\ldots\right.$ to $\frac{\mathrm{n}}{2}$ terms $)+\left(1+3+\ldots\right.$ to $\frac{\mathrm{n}}{2}$ terms $)$

Sum of a G.P. series is represented by the formula $\mathrm{S}_{\mathrm{n}}=\mathrm{a} \frac{\mathrm{r}^{\mathrm{n}}-1}{\mathrm{r}-1}$ 

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 = 2, 1

r = (ratio between the n term and n-1 term) 2, 3

$\frac{\mathrm{n}}{2}$ terms

$S_{n}=2 \times \frac{2^{\frac{n}{2}}-1}{2-1}+1 \times \frac{3^{\frac{n}{2}}-1}{3-1}$

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

$\Rightarrow S_{n}=2^{\frac{n}{2}+1}-2+\frac{3^{\frac{n}{2}}-1}{2}$

(iii) We can rewrite the given expression as

$\left(5^{1}+5^{2}+5^{3}+\ldots\right.$ to 8 terms $)$

Sum of a G.P. series is represented by the formula $S_{n}=a \frac{r^{n}-1}{r-1}$ 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 = 5

r = (ratio between the n term and n-1 term) 5

n = 8 terms

$\mathrm{S}_{\mathrm{n}}=5 \times \frac{5^{8}-1}{5-1}$

$\Rightarrow \mathrm{S}_{\mathrm{n}}=5 \times \frac{390625-1}{4}$

$\Rightarrow \mathrm{S}_{\mathrm{n}}=5 \times \frac{390624}{4}$

$\Rightarrow \mathrm{S}_{\mathrm{n}}=488280$