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Question

If(1+x)n=nr=0Crxr, then the value of C0C2+C4C6+C8C10+... equals

A
2n2cosnx4
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B
C1C3+C5C7+....
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C
C0+C4+C8+C12+....
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D
2n2sinnx4
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Solution

The correct option is B 2n2cosnx4
(1+x)n=nC0+nC1x+nC2x2+...nCnxn
And
(1x)n=nC0nC1x+nC2x2+...(1)nnCnxn
Adding, we get
(1+x)n+(1x)n=2[nC0+nC2x2+nC4x4+nC6x6+...]
Substituting, x=1=i we get
(1+i)n+(1i)n=2[nC0nC2+nC4x4nC6x6+...]
2n21[einπ4+einπ4]=nC0nC2+nC4x4nC6x6+...
Hence
Solving the LHS, we get 2n2cos(nπ4)

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