Question

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Answer 1

For any whole number $\sf{k,}$

• $\sf{i^{4k+1}=i}$

• $\sf{i^{4k+2}=i^2}$

• $\sf{i^{4k+3}=i^3}$

• $\sf{i^{4k+4}=i^4}$

If we add these four numbers,

$\longrightarrow\sf{i^{4k+1}+i^{4k+2}+i^{4k+3}+i^{4k+4}=i+i^2+i^3+i^4}$

$\longrightarrow\sf{i^{4k+1}+i^{4k+2}+i^{4k+3}+i^{4k+4}=i-1-i+1}$

$\longrightarrow\sf{i^{4k+1}+i^{4k+2}+i^{4k+3}+i^{4k+4}=0}$

Replacing $\sf{4k=n,}$

$\longrightarrow\sf{i^{n+1}+i^{n+2}+i^{n+3}+i^{n+4}=0}$

That is, the sum of any four consecutive powers of $\sf{i}$ is always zero.

Hence,

$\begin{aligned}\longrightarrow\ \ &\sf{i^n+i^{n+1}+i^{n+2}+i^{n+3}+i^{n+4}+i^{n+5}+i^{n+6}+i^{n+7}}\\=\ \ &\sf{\left(i^n+i^{n+1}+i^{n+2}+i^{n+3}\right)+\left(i^{n+4}+i^{n+5}+i^{n+6}+i^{n+7}\right)}\\=\ \ &\sf{0+0}\\=\ \ &\sf{0}\end{aligned}$

Hence Proved!

Answer 2

Answer:

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Step-by-step explanation:

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