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sum(r=0)^n(-1)^r .^n Cr[1/(2^r)+3/(2^(2r...

`sum_(r=0)^n(-1)^r .^n C_r[1/(2^r)+3/(2^(2r))+7/(2^(3r))+(15)/(2^(4r))+ .....mt e r m s]= (2^(m n)-1)/(2^(m n)(2^n-1))`

A

-6

B

-3

C

3

D

Cannot be determined

Text Solution

Verified by Experts

The correct Answer is:
d
`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 .
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