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Question: The sum of *n* terms of the following series \(1 + ( 1 + x ) + \left( 1 + x + x ^ { 2 } \right) + ...

The sum of n terms of the following series

1+(1+x)+(1+x+x2)+…1 + ( 1 + x ) + \left( 1 + x + x ^ { 2 } \right) + \ldots will be

A

1−xn1−x\frac { 1 - x ^ { n } } { 1 - x }

B

x(1−xn)1−x\frac { x \left( 1 - x ^ { n } \right) } { 1 - x }

C

n(1−x)−x(1−xn)(1−x)2\frac { n ( 1 - x ) - x \left( 1 - x ^ { n } \right) } { ( 1 - x ) ^ { 2 } }

D

None of these

Answer

n(1−x)−x(1−xn)(1−x)2\frac { n ( 1 - x ) - x \left( 1 - x ^ { n } \right) } { ( 1 - x ) ^ { 2 } }

Explanation

Solution

\begin{tabular} { l } S=1+(1+x)+(1+x+x2)+…….S = 1 + ( 1 + x ) + \left( 1 + x + x ^ { 2 } \right) + \ldots \ldots .. \ S=1+(1+x)+……S = \quad 1 + ( 1 + x ) + \ldots \ldots. \ \hline 0=(1+x+x2+….0 = \left( 1 + x + x ^ { 2 } + \ldots . \right.. to nn terms) −Tn- T _ { n } \quad (on subtracting) \end{tabular}

∴ Tn=1−xn1−xT _ { n } = \frac { 1 - x ^ { n } } { 1 - x }

Sn=∑n=1nTn=∑n=1n1−xn1−xS _ { n } = \sum _ { n = 1 } ^ { n } T _ { n } = \sum _ { n = 1 } ^ { n } \frac { 1 - x ^ { n } } { 1 - x } =11−x∑n=1n1−11−x∑n=1nxn=11−x⋅n−11−x⋅x⋅(1−xn1−x)= \frac { 1 } { 1 - x } \sum _ { n = 1 } ^ { n } 1 - \frac { 1 } { 1 - x } \sum _ { n = 1 } ^ { n } x ^ { n } = \frac { 1 } { 1 - x } \cdot n - \frac { 1 } { 1 - x } \cdot x \cdot \left( \frac { 1 - x ^ { n } } { 1 - x } \right)

=n1−x−x(1−xn)(1−x)2=n(1−x)−x(1−xn)(1−x)2= \frac { n } { 1 - x } - \frac { x \left( 1 - x ^ { n } \right) } { ( 1 - x ) ^ { 2 } } = \frac { n ( 1 - x ) - x \left( 1 - x ^ { n } \right) } { ( 1 - x ) ^ { 2 } }