Question

Sum to first n terms :
$3,33,303,3003,.....$

Answer 1

$3,33,303,3003,....$

$3,(3\cdot 10+3),(3\cdot 100+3),(3\cdot 1000+3).....$

$Generalterm{a}_n=3\cdot {10}^{n-1}+3=3(\cdot {10}^{n-1}+1)$

$\therefore S_n=\sum 3({10}^{n-1}+1)$

$=3\sum ({10}^{n-1}+1)$

$=3[\sum {10}^{n-1}+\sum 1]$

$=3[1\cdot (\frac{1-{10}^n}{1-10})+n]$

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

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