Binomial Theorem 2 Question 16

19. Prove that C122C22+3C322nC2n2=(1)nnCn

(1979, 4M)

Show Answer

Solution:

  1. We know that, (1+x)2n=C0+C1x+C2x2++C2nx2n On differentiating both sides w.r.t. x, we get 2n(1+x)2n1=C1+2C2x+3C3x2

++2nC2nx2n1

and 11x2n=C0C11x+C21x2C31x3

++C2n1x2n

On multiplying Eqs. (i) and (ii), we get

2n(1+x)2n111x2n=[C1+2C2x+3C3x2++2nC2nx2n1]×C0C11x+C21x2..+C2n1x2n

Coefficient of 1x on the LHS

= Coefficient of 1x in 2n1x2n(1+x)2n1(x1)2n= Coefficient of x2n1 in 2n(1x2)2n1(1x)=2n(1)n1(2n1)Cn1(1)=(1)n(2n)(2n1)!(n1)!n!=(1)nn(2n)!(n!)2n=(1)nnCn

Again, the coefficient of 1x on the RHS

=(C122C22+3C322nC2n2)

From Eqs. (iii) and (iv),

C122C22+3C322nC2n2=(1)nnCn



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक