# Find the sum of series 4 + 44 + 444 + 4444 +------upto n terms.

$\begin{array}{1 1} \infty \\ \frac{4}{9}[\frac{10(10^n-1)}{9}-n] \\44^4-4 \\ \frac{4(4^n-1)}{3}\end{array}$

Answer : (b) $\frac{4}{9}[\frac{10(10^n-1)}{9}-n]$
Explanation: S=4+44+444+-------n terms
$=4(1+11+111+1111+------n\;terms)$
$=\frac{4}{9}(9+99+999+-----------n\;terms)$
$=\frac{4}{9}(10-1+100-1+1000-1+-----------n\;terms)$
$=\frac{4}{9}((10-1)+(10^2-1)+(10^3-1)+-----------n\;terms)$
$=\frac{4}{9}[10+10^2+10^3+-----------n\;terms-n]$
$=\frac{4}{9}[\frac{10(10^n-1)}{10-1}-n]$
$=\frac{4}{9}[\frac{10(10^n-1)}{9}-n].$