SOLUTION: 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)

Algebra ->  Sequences-and-series -> SOLUTION: 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)      Log On


   

Question 1199617: 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)
Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.
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,

                       %28n%2A%28n%2B1%29%29%2F2.   



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  %28n%2A%28n%2B1%29%29%2F2.

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 . . .