Given positive integers `r >1,n >2` and that the coefficient of `(3r d)t ha n d(r+2)t h` terms in the binomial expansion of `(1+x)^(2n)` are equal. Then `n=2r` b. `n=2r+1` c. `n=3r` d. none of these

Options

A. 2r-1
B. 2r
C. 2r+1
D. 2r+2

Answer

c

Explanation

Clearly,
` L m= 2n xx""^(2n-1)C_(3r-1) and M = 2n""^(2n-1)C_(r+1)`
` therefore (r +2) L = 3r (M)`
`rArr (r+2) 2nxx""^(2n-1)C_(3r -1) = 3rxx2n ""^(2n-1) C_(r+1)`
`rArr ((2n-1)!(r+2))/((3r -1)!(2n -3r)) = (3rxx(2n -1))/((2n - r - 2)!(r+1)!)`
`rArr (1)/(3r!(2n -3r)!) = (1)/((r + 2 )!(2n - r - 2)1)`
`rArr ((2n)!)/(3r!(2n -3r)!) ((2n)!)/((r+2)!(2n -r-2)!)`
`rArr ""^(2n)C_(3r) = ""^(2n)C_(r+2)`
`rArr 3r + r + 2 2n rArr n = 2r +1` .