document.write( "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.
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Algebra.Com's Answer #833563 by ikleyn(52790)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "Resolve the following sum: ∑ (𝑘 + 𝑖^𝑘), from k=1 to n=4q where q is a natural number and 𝑖^2 = −1.
\n" ); document.write( "A) 𝑛(2𝑛 + 1) B) 2𝑛(4𝑛 + 1) C) 0 D) 𝑛(4𝑛 + 1) E) 2𝑛(4𝑛 − 1)
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document.write( "In this sum, ∑ (𝑘 + 𝑖^𝑘), from k=1 to n=4q, you can group the addends ∑ (𝑖^𝑘) separately in q groups, \r\n" );
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document.write( "where each group of 4 addends is repeating  (i + i^2 + i^3 + i^4) = (i - 1 - i + 1) = 0.\r\n" );
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document.write( "So,  ALL  THESE  ADDENDS  with degrees of \"i\" will cancel / annihilate each other \r\n" );
document.write( "just inside of each group of four consecutive addends, and will contribute 0 (zero) to the final sum.\r\n" );
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document.write( "Thus the final sum will be  ∑ 𝑘  from k=1 to n,  which is WELL KNOWN sum of the first n natural numbers,\r\n" );
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document.write( "                       \"%28n%2A%28n%2B1%29%29%2F2\".   \r\n" );
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document.write( "ANSWER.  The sum: ∑ (𝑘 + 𝑖^𝑘), from k=1 to n=4q, where q is a natural number and 𝑖^2 = −1,\r\n" );
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document.write( "         is equal to  \"%28n%2A%28n%2B1%29%29%2F2\".\r\n" );
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\n" ); document.write( "\n" ); document.write( "        The correct answer is  NOT  in your list of answers;  so your problem's formulation, \r
\n" ); document.write( "\n" ); document.write( "            as it is presented in the post,  is   D E F E C T I V E,  unfortunately.\r
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\n" ); document.write( "\n" ); document.write( "It is,  probably,  one million thousandth case
\n" ); document.write( "when I see incorrect problem formulation posted to this forum,\r
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\n" ); document.write( "\n" ); document.write( "so  I  am not surprised anymore . . . \r
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