Question

Find the sum to $n$ terms of the series
$6+66+666+........$

Answer 1

Given series is $6+66+666+......$ $n$ terms
$6(1+11+111+.....nterms)$
$\frac{6}{9}(9+99+999+......+nterms)$
$=\frac{2}{3}\left[\right(10-1)+(100-1)+(1000-1)+......+nterms)$
$=\frac{2}{3}(10+100+1000+.....nterms-n)$
$=\frac{2}{3}(\frac{10({10}^n-1)}{10-1}-n)$
Thus, $a=10,r=10$