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Resolve the following sum: ∑ (𝑘 + 𝑖^𝑘), from k=1 to n=4q where q is a natural number and 𝑖^2 = −1.
A) 𝑛(2𝑛 + 1) B) 2𝑛(4𝑛 + 1) C) 0 D) 𝑛(4𝑛 + 1) E) 2𝑛(4𝑛 − 1)
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In this sum, ∑ (𝑘 + 𝑖^𝑘), from k=1 to n=4q, you can group the addends ∑ (𝑖^𝑘) separately in q groups,
where each group of 4 addends is repeating (i + i^2 + i^3 + i^4) = (i - 1 - i + 1) = 0.
So, ALL THESE ADDENDS with degrees of "i" will cancel / annihilate each other
just inside of each group of four consecutive addends, and will contribute 0 (zero) to the final sum.
Thus the final sum will be ∑ 𝑘 from k=1 to n, which is WELL KNOWN sum of the first n natural numbers,
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ANSWER. The sum: ∑ (𝑘 + 𝑖^𝑘), from k=1 to n=4q, where q is a natural number and 𝑖^2 = −1,
is equal to .
Solved.
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The correct answer is NOT in your list of answers; so your problem's formulation,
as it is presented in the post, is D E F E C T I V E, unfortunately.
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It is, probably, one million thousandth case
when I see incorrect problem formulation posted to this forum,
so I am not surprised anymore . . .