Question

Sum of the series $\sum _{r=0}^n(-1{)}^{r{}^n}{C}_r[i^{5r}+i^{6r}+i^{7r}+i^{8r}]$, is

• A. $2^n$
• B. $2^{\frac{n}{2}+1}$
• C. $n^n+2^{\frac{n}{2}+1}$
• D. $2^n+2^{\frac{n}{2+1}}\cos \frac{n\pi }{4}$

Answer 1

The correct option is D $2^n+2^{\frac{n}{2+1}}\cos \frac{n\pi }{4}$

${\sum }_{r=0}^n{(-1)}^r{{}^{n}C}_r[i^{5r}+i^{6r}+i^{7r}+i^{8r}]$

Consider the given equation.

In order to compute $i^n$ for $n>4,$ we divide $n$ by $4$ and obtain the reminder $r.$

Let $m$ be the quotient when $n$ is divided by $4$

$\Rightarrow n=4m+r$ where $0\le r\le 4$

From the equation for $i^{5r}$,

$\Rightarrow i^{5r}={\left(i^4\right)}^m.i^{5r}=i^{4m+5r}=i^{5n}$

Similarly for $i^{6r},i^{7r}\xi i^{8r}\Rightarrow i^{6r},i^{7r},i^{8r}$

$\therefore {\sum }_{r=0}^n{(-1)}^r{{}^{n}C}_r[i^{5n}+i^{6n}+i^{7n}+i^{8n}]$

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