Question

Find sum of $3+33+333+3333+$....... up to n terms.

Answer 1

From the given series, take common factor 3

Thus

$S=3(1+11+111+1111+...............nthterm)$

Or

$S=3\{1+(1+10)+(1+10+100)+(1+10+100+1000)+...tonterms\}$

Thus the nth term.

$t_n=1+10+100+..............+{10}^{n-1}$

This is a G.P. where first term is 1 and common ratio is 10 ,

Thus sum of n terms is

$t_n=\frac{1(1-{10}^n)}{1-10}{t}_n=\frac{({10}^n-1)}{9}$

Or

Thus

$S=3{\sum }_{i=1}^n{t}_n=3{\sum }_{i=1}^n\frac{{10}^n-1}{9}$

Hence the sum,

$S=3\sum \frac{{10}^n}{9}-3\sum \frac{1}{9}$

Or

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

Or

$S=3\{10.\frac{({10}^n-1)}{81}-\frac{n}{9}\}$

Thus,

$S=\{10.\frac{({10}^n-1)}{27}-\frac{n}{3}\}=\frac{{10}^{n+1}-10-9n}{27}$