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Q.

Let a0,a1,a2anR be in an arithmetic progression and let 
C0,C1,C2Cn be binomial coefficients. Then k=0nak.Ck=
 

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a

a0+an.2n-1

b

12a0+an

c

a0+an

d

0

answer is B.

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Detailed Solution

S=a0 C0+a1 C1+a2 C2+.....+an CnS=a0 Cn+a1 Cn-1+a2 Cn-2+.....+an C02S=a0+an C0+a1+an-1 C1+a2+an-2 C2  Since a0, a1, a2 ...... an are in A.P. a0+an=a1+an-1=an+an-2=......2S=a0+an C0+C1+C2+......Cn2S=a0+an 2nS=a0+an 2n-1

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