The sum `sum_(k=1)^(20) k (1)/(2^(k))` is equal to

Options

A. `2 - (11)/(2^(19))`
B. `1 - (11)/(2^(20))`
C. `2 - (3)/(2^(17))`
D. `2 - (21)/(2^(20))`

Answer

A

Explanation

Let `S = underset(h =1)overset(20)sum k ((1)/(2^(k)))`
`S = (1)/(2) + (2)/(2^(2)) + (3)/(2^(3)) + (4)/(2^(4)) + ...+ (20)/(2^(20))`...(i)
On multiplying by `((1)/(2))` both sides, we get
`(S)/(2) = (1)/(2^(2)) + (2)/(2^(3)) + (3)/(2^(4)) + ...+ (19)/(2^(20)) + (20)/(2^(21))`...(ii)
On subtracting Eq. (ii) from Eq. (i), we get
`S - (S)/(2) = (1)/(2) + (1)/(2^(2)) + (1)/(2^(3)) + ...+ (1)/(2^(20)) - (20)/(2^(21))`
`rArr (S)/(2) = ((1)/(2) (1 - (1)/(2^(20))))/(1 - (1)/(2)) - (20)/(2^(21))" " [ :' " sum of "GP = (a(1- r^(n)))/(1 -r) , r lt 1]`
`(S)/(2) = 1 - (1)/(2^(20)) - (20)/(2^(21)) = 1 - (1)/(2^(20)) - (10)/(2^(20)) = 1 - (11)/(2^(20))`
`rArr S = 2 - (11)/(2^(19))`