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Question

C0C2+C1C3+C2C4+....+Cn−2Cn=

A
2nCn2
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B
2nCn
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C
2nCn1
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D
2nC2n2
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Solution

The correct option is A 2nCn2
(1+x)n=Co+C1x+C2x2+.....+Cnxn ………….(1)
(x+1)n=Coxn+C1xn1+.....+Cnxo ……………(2)
Multiplying (1) & (2)
(Co+C1x+C2x2+....+Cnxn)×(Coxn+C1xn1+....Cnxo)=(1+x)2n
Comparing the coefficients of xn2 or xn+2 both sides.
We get
CoC2+C1C3+C2C4+....+Cn1Cn
=2nCn2 or =2nCn+1.

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