Question

Find the sum of the following finite series
(i) $1+0.1+0.01+0.001+.....+(0.1{)}^9$
(ii) $1+11+111+....$ to $20$ terms

Answer 1

(i) It is a G.P. series

Sum of $n$ terms of G.P series $=a\left(\frac{r^n-1}{r-1}\right)$

where $a$ is the first term and $r$ is the common ratio.

So sum of the given series with $a=1$ and $r=0.1$

is $=1\left(\frac{{0.1}^{10}-1}{0.1-1}\right)=\mathrm{1.111...}$ upto $10$ digits

In (ii)

Let $L$ be the required sum

$L=1+11+\mathrm{111...}$to $20$ terms

$L=d\frac{1}{9}(9+99+\mathrm{999...}$ to $20$ terms$)$

$L=\frac{1}{9}(10-1+{10}^2-1+{10}^3-\mathrm{1...}$ upto $20$ terms$)$

$L=\frac{1}{9}(10+{10}^2+{10}^3...+{10}^{20}-20)$

$L=\frac{1}{9}(10\left(\frac{{10}^{20}-1}{10-1}\right])-20)=12345679012345679010$