Sum of the series ∑r=0n-1r nCr i5r+i6r+i7r+i8r

∑r=0n-1r nCr i5r+i6r+i7r+i8r =∑r=0n (-1)r Cr n ir+i2r+i3r+1 =∑r=0n (-1)r Cr n ir+∑r=0n Cr n (-1)r (i2)r+∑r=0n (-1)r Cr n (i3)r+∑r=0n (-1)r Cr n =(1-i)n+(1-i2)n+(1-i3)n+1-1n =(1-i)n+(1+1)n+(1+i)n+1-1n =2n+2n/2cosπ4-i sinπ4n+2n/2cosπ4+i sinπ4n =2n+2n/2cos nπ4-i sin nπ4+cos nπ4+i sin nπ4 =2n+2n2·2cosnπ4 =2n+2n2+1 cosnπ4

Sum of the series ∑r=0n-1r nCr i5r+i6r+i7r+i8r

∑r=0n-1r nCr i5r+i6r+i7r+i8r =∑r=0n (-1)r Cr n ir+i2r+i3r+1 =∑r=0n (-1)r Cr n ir+∑r=0n Cr n (-1)r (i2)r+∑r=0n (-1)r Cr n (i3)r+∑r=0n (-1)r Cr n =(1-i)n+(1-i2)n+(1-i3)n+1-1n =(1-i)n+(1+1)n+(1+i)n+1-1n =2n+2n/2cosπ4-i sinπ4n+2n/2cosπ4+i sinπ4n =2n+2n/2cos nπ4-i sin nπ4+cos nπ4+i sin nπ4 =2n+2n2·2cosnπ4 =2n+2n2+1 cosnπ4

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

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$2^{n}$

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

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

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

answer is D.

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Detailed Solution

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

$\begin{array}{rcl} = & \sum_{r = 0}^{n} {(- 1)}^{r} {}^{n}C_{r} (i^{r} + i^{2 r} + i^{3 r} + 1) \end{array}$

$\begin{array}{rcl} = & \sum_{r = 0}^{n} {(- 1)}^{r} {}^{n}C_{r} i^{r} + \sum_{r = 0}^{n} {}^{n}C_{r} {(- 1)}^{r} {(i^{2})}^{r} + \sum_{r = 0}^{n} {(- 1)}^{r} {}^{n}C_{r} {(i^{3})}^{r} + \sum_{r = 0}^{n} {(- 1)}^{r} {}^{n}C_{r} \end{array}$

$\begin{array}{rcl} = & {(1 - i)}^{n} + {(1 - i^{2})}^{n} + {(1 - i^{3})}^{n} + {(1 - 1)}^{n} \end{array}$

$\begin{array}{rcl} = & {(1 - i)}^{n} + {(1 + 1)}^{n} + {(1 + i)}^{n} + {(1 - 1)}^{n} \end{array}$

$\begin{array}{rcl} = & 2^{n} + 2^{n / 2} {(\cos \frac{π}{4} - i \sin \frac{π}{4})}^{n} + 2^{n / 2} {(\cos \frac{π}{4} + i \sin \frac{π}{4})}^{n} \end{array}$