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Question

Find the value of nC1+nC5+nC9+nC13.....

A
2n2(1+sinnπ4)
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B
2n22sinnπ4+2n2
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C
2n12sinnπ4+2n2
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D
2n2sinnπ4+2n1
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Solution

The correct option is B 2n22sinnπ4+2n2
we have (1+x)n=nC0+nC1x+nC2x2+nC3x3+nC4x4+....
Putting x = - 1;
2n=nC0+nC1+nC2+nC3+nC4.... (1)
Putting x = - 1;
0=nC0nC1+nC2nC3+nC4.... (2)
subtracting (1)from (2); we get
2n1=nC1+nC3+nC5+nC7+nC9+... (3)
Putting x = i;
(1+i)n=nC0+nC1i+nC2i2+nC3i3+nC4i4+...2n2(12+i12)n=nC0+nC1i+nC2(1)+nC3(i)+nC4i+...2n2(cosπ4+i sinπ4)n=(nC0nC2+nC4...)+i(nC1nC3++C5....)2n2eiπn4=(nC0nC2+nC4...)+i(nC1nC3++C5...)2n2(cosnπ4+i sinnπ4)=(nC0nC2+nC4...)+i(nC1nC3++C5....)
Comparing the inaginary parts;
2n2sinnπ4=(nC1nC3++C5....)(4)
Adding (3)and (4) ; we get
nC1+nC5+nC9+nC13....=2n22sinnπ4+2n2

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