If `sum_(r=0)^(n){("^(n)C_(r-1))/('^(n)C_(r )+^(n)C_(r-1))}^(3)=(25)/(24)`, then `n` is equal to

Options

A. `3`
B. `4`
C. `5`
D. `6`

Answer

C

Explanation

`(c )` Let `t_(r )=('^(n)C_(r-1))/('^(n)C_(r )+^(n)C_(r-1))=(1)/(('^(n)C_(r ))/('^(n)C_(r-1))+1)=(1)/((n-r+1)/(r )+1)`
`:.t_(r )=(r )/(n+1)`
Now,
`S=sum_(r=0)^(n){t_(r )}^(3)`
`impliesS=sum_(r=0)^(n)(r^(3))/((n+1)^(3))=(1)/((n+1)^(3))sum_(r=0)^(n)r^(3)`
`impliesS=(1)/((n+1)^(3)){(n(n+1))/(2)}^(2)impliesS=(n^(2))/(4(n+1))`
Now, `S=(25)/(24)` (given) `:.n=5`