First slide
Binomial theorem for positive integral Index
Question

Let m be the smallest positive integer such that the coefficient of X2 in the expansion of (1+x)2+(1+x)3++(1+x)49+(1+mx)50 is (3n+1) 51C3 for some positive integer n. Then the value of n is _________.

Moderate
Solution

Coefficient of x2 in expansion

=1+3C2+4C2+5C2++49C2+50C2m2       as nCr+nCr1=n+1Cr

= 3C3+3C2+4C2+5C2++49C2+50C2m2= 4C3+4C2++50C2m2=5C3++49C2+50C2m2=50C3+50C2m2+50C250C2=51C3+50C2m21=(3n+1) 51C3( given )

 3n51350C2=50C2m21

m2151=nm2-1=51nm2=51n+1

n=5m2=256

value of n is 5.

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