Sum of the series ∑r=0n(−1)r nCri5r+i6r+i7r+i8r is

∑r=0n(−1)r nCri5r+i6r+i7r+i8r=∑r=0n(−1)r nCrir+i2r+i3r+1=∑r=0n(−1)r nCrir+∑r=0n(−1)r nCr(i2)r+∑r=0n(−1)r nCr(i3)r+∑r=0n(−1)r nCr=(1−i)n+(1−i2)n+(1−i3)+(1−1)n=(1−i)n+2n+(1+i)n=2n+2n2cosπ4−isinπ4n+2n2cosπ4+isinπ4n=2n+2n2+1cosnπ4Therefore, the value of the given expression is 2n+2n2+1cosnπ4

Sum of the series ∑r=0n(−1)r nCri5r+i6r+i7r+i8r is

∑r=0n(−1)r nCri5r+i6r+i7r+i8r=∑r=0n(−1)r nCrir+i2r+i3r+1=∑r=0n(−1)r nCrir+∑r=0n(−1)r nCr(i2)r+∑r=0n(−1)r nCr(i3)r+∑r=0n(−1)r nCr=(1−i)n+(1−i2)n+(1−i3)+(1−1)n=(1−i)n+2n+(1+i)n=2n+2n2cosπ4−isinπ4n+2n2cosπ4+isinπ4n=2n+2n2+1cosnπ4Therefore, the value of the given expression is 2n+2n2+1cosnπ4

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Sum of the series $\sum_{r = 0}^{n} {(- 1)}^{r}^{n} C_{r} [i^{5 r} + i^{6 r} + i^{7 r} + i^{8 r}]$ is

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$2^{n}$

$2^{\frac{n}{2} + 1} \cos \frac{n π}{4}$

$2^{n} + 2^{\frac{n}{2} + 1}$

$2^{n} + 2^{\frac{n}{2} + 1} \cos \frac{n π}{4}$

answer is D.

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Detailed Solution

$\sum_{r = 0}^{n} {(- 1)}^{r}^{n} C_{r} [i^{5 r} + i^{6 r} + i^{7 r} + i^{8 r}]$ $= \sum_{r = 0}^{n} {(- 1)}^{r}^{n} C_{r} [i^{r} + i^{2 r} + i^{3 r} + 1]$

$= \sum_{r = 0}^{n} {(- 1)}^{r}^{n} C_{r} i^{r} + \sum_{r = 0}^{n} {(- 1)}^{r}^{n} C_{r} {(i^{2})}^{r} + \sum_{r = 0}^{n} {(- 1)}^{r}^{n} C_{r} {(i^{3})}^{r} + \sum_{r = 0}^{n} {(- 1)}^{r}^{n} C_{r}$

$= {(1 - i)}^{n} + {(1 - i^{2})}^{n} + (1 - i^{3}) + {(1 - 1)}^{n}$

$= {(1 - i)}^{n} + 2^{n} + {(1 + i)}^{n}$

$= 2^{n} + 2^{\frac{n}{2}} {\{\cos \frac{π}{4} - i \sin \frac{π}{4}\}}^{n} + 2^{\frac{n}{2}} {\{\cos \frac{π}{4} + i \sin \frac{π}{4}\}}^{n} = 2^{n} + 2^{\frac{n}{2} + 1} \cos \frac{n π}{4}$