Find the sum of the series
`sum_(r=1)^(n) (-1)^(r)""^(n)C_(r) {(1)/(2^(r)) + (3^(r))/(2^(2r)) + (7^(r))/(2^(3r)) + (15^(r))/(2^(4r)) + ..."upto m terms"}` .

Options

A. -6
B. -3
C. 3
D. Cannot be determined

Answer

d

Explanation

`because sum_(r=0)^(n) (-1)^(r) ""^(n)C_(r)[((1)/(2))^(r) + ((3)/(4))^(r) + ((7)/(8))^(r) +..."upto m terms"]`
` = (1 - (1)/(2))^(n) + (1 - (3)/(4))^(n) + (1 - (7)/(8))^(n) +... "upto m terms" ]`
` (1)/(2^(n)) + (1)/(2^(2n)) + (1)/(2^(3n)) + ..." upto m terms" `
` = ((1)/(2^(n))[1-((1)/(2^(n)))^(m)])/((1- (1)/(2^(n))) )= ((1)/(2^(n)-1))(1-(1)/(2^(mn)))`
` therefore f (n) = (1)/(2^(n) -1)`
` therefore int_(-3)^(3) f(x^(3) " In x) * dt " (x^(3) ` in x)
`=int _(-3) ^(3) (1)/((2^(x^(3)"In x")-1))*(3x^(2) " In " x + x^(2))dx`
Since , In x cannot be defined for ` x lt 0 `
` therefore ` Above integral cannot be calculated .