Let m be the smallest positive integer such that the coefficient of x 2 in the expansion of 1 + x 2 + 1 + x…

Let m be the smallest positive integer such that the coefficient of x2 in the expansion of 1+x2+1+x3++1+x49+1+mx50 is 3n+1 51C3 for some positive integer n.
Then the value of n is

Solution

Coefficient of x2 in the expansion of
1+x2+1+x3+1+x49+1+mx50 is
2C2+3C2+49C2+50C2m2=3n+151C3
3C3+3C2+49C2+50C2m2=3n+151C3     (Use n C r + n C r+1 = n+1 C r+1 )
50C3+50C2m2=3n+1 51C3
50.49.486+50.492 m2=3n+151.50.496
m2=51n+1 must be a perfect square
By trial ⇒ n=5 and m=16     (M,nN)
n =5

Asked in: JEE Advanced 2016 (Paper 1)

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