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Question: The sum to n terms of the series, 1 + \(\left( 1 + \frac { 1 } { 2 } + \frac { 1 } { 2 ^ { 2 } } \ri...

The sum to n terms of the series, 1 + (1+12+122)\left( 1 + \frac { 1 } { 2 } + \frac { 1 } { 2 ^ { 2 } } \right)+

(1+12+122+123+124)\left( 1 + \frac { 1 } { 2 } + \frac { 1 } { 2 ^ { 2 } } + \frac { 1 } { 2 ^ { 3 } } + \frac { 1 } { 2 ^ { 4 } } \right) + …… is

A

2n – 43\frac { 4 } { 3 } {1122n}\left\{ 1 - \frac { 1 } { 2 ^ { 2 n } } \right\}

B

2n + 43\frac { 4 } { 3 } {1122n}\left\{ 1 - \frac { 1 } { 2 ^ { 2 n } } \right\}

C

2n + 43\frac { 4 } { 3 } {1+122n}\left\{ 1 + \frac { 1 } { 2 ^ { 2 n } } \right\}

D

None of these

Answer

None of these

Explanation

Solution

Sn = 1 + 1(12)3112\frac { 1 - \left( \frac { 1 } { 2 } \right) ^ { 3 } } { 1 - \frac { 1 } { 2 } } + 1(12)5112\frac { 1 - \left( \frac { 1 } { 2 } \right) ^ { 5 } } { 1 - \frac { 1 } { 2 } } + ….. n terms

= 1 + 2 (1(12)3)\left( 1 - \left( \frac { 1 } { 2 } \right) ^ { 3 } \right)+ 2 (1(12)5)\left( 1 - \left( \frac { 1 } { 2 } \right) ^ { 5 } \right)+ ……. n times

= 1 + (2 + 2 + …. (n – 1) times) –

[(12)2+(12)4+..(n1)\left[ \left( \frac { 1 } { 2 } \right) ^ { 2 } + \left( \frac { 1 } { 2 } \right) ^ { 4 } + \ldots . . ( \mathrm { n } - 1 ) \right. times ]]

= 1 + 2(n –1) – (1/2)21(12)2\frac { ( 1 / 2 ) ^ { 2 } } { 1 - \left( \frac { 1 } { 2 } \right) ^ { 2 } } (1(12)n1)\left( 1 - \left( \frac { 1 } { 2 } \right) ^ { \mathrm { n } - 1 } \right)

= 2n –1 – 13\frac { 1 } { 3 } + 23.22n\frac { 2 } { 3.2 ^ { 2 \mathrm { n } } } = 2n –43\frac { 4 } { 3 } (1+122n)\left( 1 + \frac { 1 } { 2 ^ { 2 n } } \right)