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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Sequence and Series

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

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

1 Answer

Answer : (a) $\;\large\frac{n(n+1)}{2}+\large\frac{1}{9}\;(1-(0.1)^n)$
Explanation : $\;S=\sum_{r=1}^{n}\;a_{r}$
$S=\sum_{r=1}^{n}\;(r+(0.1)^r)$
$=\large\frac{n(n+1)}{2}+\large\frac{(0.1)(1-(0.1)^n)}{(1-0.1)}$
$=\large\frac{n(n+1)}{2}+\large\frac{1}{9}\;(1-(0.1)^n)\;.$
answered Jan 23, 2014 by yamini.v
 

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