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sum(r=0)^(n-1) (""^(n)C(r))/(""^(n)C(r) ...

`sum_(r=0)^(n-1) (""^(n)C_(r))/(""^(n)C_(r) + ""^(n)C_(r+1))` is equal to

A

`(n-1)/(n+1)`

B

`(n+1)/(2)`

C

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

D

`(n)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
d
We have
` sum_(r=0)^(n-1) (""^(n)C_(r))/(""^(n)C_(r) + ""^(n)C_(r+1)) `
` sum_(r=0)^(n-1) (""^(n)C_(r))/( ""^(n+1)C_(r+1)) `
` sum_(r=0)^(n-1) (""^(n)C_(r))/( (n+1)/(r +1)""^(C_(r))) `
` sum_(r=0)^(n-1) (r+1)/(r+1)=(1)/(n+1)sum_(r=0)^(n-1) ( r+1) = (1)/(n+1) xx(n (n+1))/(n +1)= (n)/(2).` .
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