Binomial Theorem 2 Question 7

10. If Cr stands for nCr, then the sum of the series

2n2!n2!n![C022C12+3C22+(1)n(n+1)Cn2]

where n is an even positive integer, is

(a) (1)n/2(n+2)

(b) (1)n(n+1)

(1986, 2M) (1997C, 5M)

Show Answer

Answer:

Correct Answer: 10. (a)

Solution:

  1. We have,

C022C12+3C224C32++(1)n(n+1)Cn2=[C02C12+C22C32++(1)nCn2][C122C22+3C32+(1)nnCn2]=(1)n/2n!n2!n2!(1)n21n2n!n2!n2!=(1)n/2n!n2!n2!1+n2

2n2!n2!n![C022C12+3C22+(1)r(n+1)Cn2]

=2n2!n2!n!(1)n/2n!n2!n2!(n+2)2=(1)n/2(n+2)



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