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Sum of the first n terms of the series 1...

Sum of the first `n` terms of the series `1/2+3/4+7/8+(15)/(16)+.........` is equals to (a).`2^n+n-1` (b). `n-1+2^(-n)` (c). `n+2^(-n)-1` (d). `2^n+1`

A

`2^(n)+n-1`

B

`1-2^(-n)`

C

`n+2^(-n)-1`

D

`2^(n)-1`

Text Solution

Verified by Experts

The correct Answer is:
C

`:.S_(n)=(1)/(2)+(3)/(4)+(7)/(8)+(15)/(16)+"…."n " up to terms "`
`=(1-(1)/(2))+(1-(1)/(4))+(1-(1)/(8))+"...."n " upto terms "`
`=n-(1)/(2)(1+(1)/(2)+(1)/(2^(2))+"...."+(1)/(2^(n-1)))`
`=n-(1)/(2)1([1-((1)/(2))^(n)])/((1-(1)/(2))) " " [" by sum GP, " S_(n)=(a(1-r^(n)))/(1-r)," if " oltrlt1]`
`n-1_(1)/(2^(n))=n-1+2^(-n)`
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