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Question: The sum to *n* terms of the series \(1 + 3 + 7 + 15 + 31 + \ldots \ldots\) is...

The sum to n terms of the series 1+3+7+15+31+1 + 3 + 7 + 15 + 31 + \ldots \ldots is

A

2n+1n2 ^ { n + 1 } - n

B

2n+1n22 ^ { n + 1 } - n - 2

C

2nn22 ^ { n } - n - 2

D

None of these

Answer

2n+1n22 ^ { n + 1 } - n - 2

Explanation

Solution

\begin{tabular} { l } S=1+3+7+15+31+.S = 1 + 3 + 7 + 15 + 31 + \ldots .. \ S=1+3+7+15+S = 1 + 3 + 7 + 15 + \ldots \ldots. \ \hline 0=(1+2+4+8+16+.0 = \left( 1 + 2 + 4 + 8 + 16 + \ldots . \right.. to nn terms) Tn- T _ { n } \quad (on subtracting) \end{tabular}

Tn=1+2+4+8+T _ { n } = 1 + 2 + 4 + 8 + \ldots \ldots \ldots \ldots. . to nn terms =12n121=2n1= 1 \cdot \frac { 2 ^ { n } - 1 } { 2 - 1 } = 2 ^ { n } - 1 Sn=n=1nTn=n=1n(2n1)=n=1n2nn=1n1=2(2n121)n=2n+1n2S _ { n } = \sum _ { n = 1 } ^ { n } T _ { n } = \sum _ { n = 1 } ^ { n } \left( 2 ^ { n } - 1 \right) = \sum _ { n = 1 } ^ { n } 2 ^ { n } - \sum _ { n = 1 } ^ { n } 1 = 2 \cdot \left( \frac { 2 ^ { n } - 1 } { 2 - 1 } \right) - n = 2 ^ { n + 1 } - n - 2