Binomial Theorem

Asked by  | 26th Jun, 2008, 06:11: PM

Expert Answer:

if we expand (1+x)n , then nC2 represents the coefficient of x2

also nCr can be written as nCn-r .

so (100C98)+(99C97)+.........+(3C1)+(2C0) =  (100C2)+(99C2)+..............+(3C2)+(2C2)

so this expression will be equal to coeficient of x2 in (1+x)100 +(1+x)99+..............+(1+x)3+(1+x)2 

sum of this series is (1+x)2( (1+x)99-1) /  ( (1+x)-1 ) = (1/x) (1+x)2( (1+x)99-1 ) = (1/x) ( (1+x)101- (1+x)2 )

so we have to find coefficient of x3 in (  (1+x)101- (1+x)2 ) . since cofficient of x3 is zero in (1+x)2

so cofficient of x2 in this expression is 101C3

Answered by  | 13th Sep, 2008, 07:38: PM

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