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Question

The value of the expression
C20C21+C22......+(1)n×C2n is

A
0, if n is odd
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B
(1)n, if n is odd
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C
(1)n/2 nCn/2, if n is even
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D
(1)n1 nCn1, if n is even
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Solution

The correct options are
B 0, if n is odd
C (1)n/2 nCn/2, if n is even
Given , C20C21+C22......(1)n×C2n
Case I : When n is odd, i.e. n=2m+1,
S=C20C21+C22.....+(1)2m×C22m+(1)2m+1×C2m+1
S=C20C21+C22.....+C22mC22m+1
=(C20C22m+1)(C21C22m)+....+(1)m(C2mC2m+1)
But
2m+1C0=2m+1C2m+1;2m+1C1=2m+1C2m etc.
Therefore, S=0
Case II: When n is even, i.e. n=2m.
S=C20C21+C22......+(1)2m×C22m
S=C20C21+C22......+C22m
Consider the coefficient of the constant term in the following expression.
[C0C1x+C2x2.....+(1)2m×C2mx2m][C0+C11x+C21x2+....+C2m1x2m]
=(1x)2m(1+1x)2m
Coefficient of x2m in (1x)2m(1+x)2m
= Coefficient of x2m in (1x2)2m
=(1)m2mCm=(1)n/2nCn/2

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