Find the sum of the series :

Solution:

The expression can be rewritten as

[Taking 8 as a common factor]

8(1+ 11 + 111+ … to n terms)

[Multiplying and dividing the expression by 9]

$=\frac{8}{9}(9+99+999+\ldots$ to $n$ terms $)$

$=\frac{8}{9}((10-1)+(100-1)+(1000-1)+\ldots$ to $n$ terms $)$

$=\frac{8}{9}((10+100+1000+\ldots$ to $n$ terms $)-(1+1+1+\ldots$ to $n$ terms $)$

$=\frac{8}{9}((10+100+1000+\ldots$ to $n$ terms $)-n)$

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 = 10

r = (ratio between the n term and n-1 term) 10

n terms

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

$\Rightarrow \mathrm{S}_{\mathrm{n}}=10 \times \frac{10^{\mathrm{n}}-1}{9}$

$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{10^{\mathrm{n}+1}-10}{9}$

∴ The sum of the given expression is

$=\frac{8}{9}((10+100+1000+\ldots$ to $n$ terms $)-n)$

$=\frac{8}{9}\left(\frac{10^{n+1}-10}{9}-n\right)$

(ii) The given expression can be rewritten as

[taking 3 common ]

= 3( 1+11+111+ …to n terms)

[Multiplying and dividing the expression by 9 ]

$=\frac{3}{9}(9+99+999+\ldots$ to $n$ terms $)$

$=\frac{3}{9}((10-1)+(100-1)+(1000-1)+\ldots$ to $n$ terms $)$

$=\frac{3}{9}((10+100+1000+\ldots$ to $n$ terms $)-(1+1+1+\ldots$ to $n$ terms $))$

$=\frac{3}{9}((10+100+1000+$ to $n$ terms $)-n)$

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 = 10

r = (ratio between the n term and n-1 term) 10

n terms

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

$\Rightarrow \mathrm{S}_{\mathrm{n}}=10 \times \frac{10^{\mathrm{n}}-1}{9}$

$\Rightarrow S_{n}=\frac{10^{n+1}-10}{9}$

∴ The sum of the given expression is

$=\frac{3}{9}((10+100+1000+$ to $n$ terms $)-n)$

$=\frac{3}{9}\left(\frac{10^{n+1}-10}{9}-n\right)$

(iii) We can rewrite the expression as

[taking 7 as a common factor]

= 7(0.1+0.11+0.111+ … to n terms)

[multiplying and dividing by 9 ]

$=\frac{7}{9}(0.9+0.99+0.999+\ldots$ to $n$ terms $)$

$=\frac{7}{9}((1-0.1)+(1-0.01)+(1-0.001)+\ldots$ to $n$ terms $)$

$=\frac{7}{9}((1+1+1+\ldots$ to $n$ terms $)-(0.1+0.01+0.001+\ldots$ to $n$ terms $))$

$=\frac{7}{9}(n-(0.1+0.01+0.001+\ldots$ to $n$ terms $))$

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

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 = 0.1

r = (ratio between the n term and n-1 term) 0.1

n terms

$\mathrm{S}_{\mathrm{n}}=0.1 \times \frac{1-0.1^{\mathrm{n}}}{1-0.1}$

$\Rightarrow \mathrm{S}_{\mathrm{n}}=0.1 \times \frac{1-0.1^{\mathrm{n}}}{0.9}$

[multiplying both numerator and denominator by 10]

$\Rightarrow S_{n}=\frac{1-0.1^{n}}{9}$

∴ The sum of the given expression is

$=\frac{7}{9}(\mathrm{n}-(0.1+0.01+0.001+\ldots$ to $\mathrm{n}$ terms $))$

$=\frac{7}{9}\left(n-\left(\frac{1-0.1^{n}}{9}\right)\right)$