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Question

If n=12m(mN),nC0nC2(2+3)2+nC4(2+3)4nC6(2+3)6+..=(1)m(p1+Q)n then PQ is:

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Solution

(1+x)n+(1x)n2=nC0+nC2x2+.....
Putting x=i2+3
S=(1+i2+3)n+(1i2+3)n2

=nC0nC2.1(2+3)2+.....
S=(2+3+i)n+(2+3+i)n2(2+3)n
=2n[(1+3+i2)n+(1+(3i2)n)]2.2n(1+3/4)n
=12(1+3/4)n[(1+eiπ/6)n+(1+eiπ/6)n]
=12×1(1+34)n×2
=1(1+34)
P=1,θ=34
PQ=14

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