1.

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

Answer»

`3`
`4`
`5`
`6`

Solution :`(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`


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