If `S_n=sum_(r=1)^n(1+2+2^2+ .......+2^r)/(2^r),` then `S_n` is equal to (a)`2^n n-1` (b) `1-1/(2^n)` (c) 2n`-1+1/(2^n)` (d) `2^n-1`

Explanation

Given that `S_n=sum_(r=1)^n(1+2+2^2+ .......+2^r)/(2^r)`

We know that,

Sum of finite terms is
,
`S_n=(a(r^n-1))/(r-1)`

here,

`a=1/(2^r)` and `r=2`

So,


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