# Evaluate sum to n terms $\;1.1+2.01+3.001+\;.....$

$(a)\;\large\frac{n(n+1)}{2}+\large\frac{1}{9}\;(1-(0.1)^n)\qquad(b)\;n+\large\frac{1}{9}\;(1-(0.1)^n)\qquad(c)\;\large\frac{n(n+1)}{2}+\large\frac{1}{10}\;(1-(0.1)^n)\qquad(d)\;\large\frac{n(n+1)}{2}+\large\frac{1}{10}\;(0.1)^n$

Answer : (a) $\;\large\frac{n(n+1)}{2}+\large\frac{1}{9}\;(1-(0.1)^n)$
Explanation : $\;S=\sum_{r=1}^{n}\;a_{r}$
$S=\sum_{r=1}^{n}\;(r+(0.1)^r)$
$=\large\frac{n(n+1)}{2}+\large\frac{(0.1)(1-(0.1)^n)}{(1-0.1)}$
$=\large\frac{n(n+1)}{2}+\large\frac{1}{9}\;(1-(0.1)^n)\;.$