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Question: The sum of \(n\) terms of the following series <img src="https://cdn.pureessence.tech/canvas_464.png...

The sum of nn terms of the following series 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

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

(1+x+x2+x3+…+xn−1)+…\left( 1 + x + x ^ { 2 } + x ^ { 3 } + \ldots + x ^ { n - 1 } \right) + \ldots

Required sum

=1(1−x){(1−x)+(1−x2)+(1−x3)\frac { 1 } { ( 1 - x ) } \left\{ ( 1 - x ) + \left( 1 - x ^ { 2 } \right) + \left( 1 - x ^ { 3 } \right) \right.

+(1−x4)+……….+ \left( 1 - x ^ { 4 } \right) + \ldots \ldots \ldots .. uptp nn terms }\}

=1(1−x)[n−{x+x2+x3+……….= \frac { 1 } { ( 1 - x ) } \left[ n - \left\{ x + x ^ { 2 } + x ^ { 3 } + \ldots \ldots \ldots . \right. \right.. upto nn terms }]\left. \} \right]

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