sum(r=0)^n (-1)^r C(n,r)((1/(2^r))+((3^r...

`sum_(r=0)^n (-1)^r C(n,r)((1/(2^r))+((3^r)/(2^(2r)))+((7^r)/(2^(3r)))+...oo` is equal to

-6

-3

Cannot be determined

Text Solution

Verified by Experts

The correct Answer is:

`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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If sum_(r=0)^(n) (-1)^(r ).""^(n)C_(r)[(1)/(2^(r))+(3^(r))/(2^(2r))+(7^(r))/(2^(3r))+"……..""to m terms"] = k(1-1/(2^(mn))) , then k =

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

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

`(1)/(2^(mn) + 1)`

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

If (1+x)^(n)=sum_(r=0)^(n)*C_(r )x^(r ) and sum_(r=0)^(n) (-1)^(r ) (C_(r ))/((r+1)^(2)=k sum_(r=0)^(n)(1)/(r+1) , then k is equal to :

`(1)/(n) `

`(1)/(n+1)`

`(n)/(n+1)`

none

underset(r=0)overset(n)(sum)(-1)^(r).^(n)C_(r)[(1)/(2^(r))+(3^(r))/(2^(2r))+(7^(r))/(2^(3r))+(15^(r))/(2^(4r))+ . . .m" terms"]=

`(2^(mn)-1)/(2^(mn)(2^(n)-1))`

`(2^(mn)-1)/(2^(n)-1)`

`(2^(mn)+1)/(2^(n)+1)`

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