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The vaule of sum(r=0)^(n-1) (""^(C(r))...

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

A

`(n)/(2) `

B

`(n+1)/(2)`

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
a
We have, `sum_(r=0)^(n-1) (""^(n)c_(r))/(""^(n)c_(r)+ ""^(n)c_(r+1))=sum_(r=1)^(n-1) (""^(n)c_(r))/(""^(n+ 1)c_(r+1))`
`= sum_(r=0)^(n-1) (""^(n)c_(r))/((n+1)/(r+1) ""^(n)c_(r)) = sum_(r=0)^(n-1) (r+1)/(n+1)`
` = (1)/(n+1)[1+ 2 + ...+ n] = (n(n+1))/(2(n+1)) = (n)/(2)`
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