Question 1199617
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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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<pre>
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,

                       {{{(n*(n+1))/2}}}.   



<U>ANSWER</U>.  The sum: ∑ (𝑘 + 𝑖^𝑘), from k=1 to n=4q, where q is a natural number and 𝑖^2 = −1,

         is equal to  {{{(n*(n+1))/2}}}.
</pre>

Solved.



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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The correct answer is &nbsp;NOT &nbsp;in your list of answers; &nbsp;so your problem's formulation, 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;as it is presented in the post, &nbsp;is  &nbsp;&nbsp;D E F E C T I V E, &nbsp;unfortunately.


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It is, &nbsp;probably, &nbsp;one million thousandth case 
when I see incorrect problem formulation posted to this forum,


so &nbsp;I &nbsp;am not surprised anymore . . .