AP EAMCET Engineering 2018 Apr 24 Shift 1 Paper

From the given series,
(1)

1

4

=(cos‌0+i‌sin‌0)

1

4

=(cos‌2‌r‌π+i‌sin‌2‌r‌π)

1

4

=cos‌

rπ

2

+i‌sin‌

rπ

2

The four roots of unity for the following equation is 1,i,−1,−i= 1,α,α2,α3
And (1+x)n=

n

∑

r=0

nCrxr
For x=1,(2)n=

n

∑

r=0

nCr
For x=α(1+α)n=

n

∑

r=0

nCrαr
For x=α2(1+α2)n=

n

∑

r=0

nCrα2r
For x=α3(1+α3)n=

n

∑

r=0

nCrα3r
Add the above equations,
2n+(1+α)+(1+α2)+(1+α3)
=

n

∑

r=0

nCr(1+αr+α2r+α3r)
Substitute the valeus in the above equation
4(nC0+nC4+nC8+........) =2n+(1+i)n+(1+i2)n+(1+i3)n
=2.2

n

2

[cos‌ ‌

nπ

4

+2

n

2

−1]
=

2 n 2 [cos‌ nπ 4 +2 n 2 −1]

2