Let m be the smallest positive integer such that the coefficient of `x^(2)` in the expansion of `(1+x)^(2)+(1+x)^(3) + "……." + (1+x)^(49) + (1+mx)^(50)` is `(3n+1) .^(51)C_(3)` for some positive integer n, then the value of n is `"_____"`.

Options

A. 16
B. 5
C. 21
D. 11

Answer

B

Explanation

We find that
`(1 + x)^(2) + (1 + x)^(3) +… (1 + x)^(49) + (1 + mx)^(50)`
` = (1 + x)^(2) { ((1 + x)^(48) -1)/((1 + x) -1)} +(1 + mx)^(50)`
` (1)/(x) {( 1 + x)^(50) - (1 + x)^(2)} + (1 +mx)^(50)`
Coeffi. Of `x^(2)` in `(1 + x)^(2) + (1 + x)^(3) + ...(1 + x)^(49) + (1 +mx)^(50)`
= Coeffi.of `x^(2)` in `[(1)/(x) {(1 +x)^(50) - (1 + x)^(2)} +(1 +mx)^(50)]`
` Coeffi. of `x^(3)` in `{ (1 + x)^(50) - (1 + x)(2)} + ` Coefficient of `x^(2)` in `(1 +mx)^(50)`
`""^(50)C_(3) + ""^(50)C_(2) m^(2)`
It is given that this coefficient is `(3n +1) ""^(51)C_(3)` .
`therefore ""^(50)C_(3) + ""^(50)C_(2) m^(2) = (3n +1)""^(51)C_(3)`
`rArr (50xx49xx48)/(3xx2xx1)+(50xx49)/(2xx1)m^(2) = (3n-1)(51xx50xx49)/(3xx2xx1)`
`rArr 16 + m^(2) = 17 (3n +1)`
`rArr m^(2) = 51n+1`
The least value of n for which `51 N+1` is a perfect square is 5 and
for this value of n the value of m is 16.
Hence,m=16 and n=5.