What is the sum of this imaginary number series problem: i+2i2+3i3+4i4+5i5+…+2002i2002?

Question

What is the sum of this imaginary number series problem:  i+2i2+3i3+4i4+5i5+…+2002i2002?

Matthew Rodriguez · Accepted Answer

The Complex multipliers can be simplified into a cyclic pattern of period 4:  i=i  i2=−1  i3=−i  i4=1  Since i4=1, we have i=i5=i9=… and i2=i6=i10=… etc. and we find:  ∑n=12002nin  =(∑k=0500(4k+1))i−(∑k=0500(4k+2))−(∑k=0499(4k+3))i+(∑k=0499(4k+4))  Each of these four sums is an arithmetic sequence, so is equal to the number of terms multiplied by the average term. The average term is equal to the average of the first and last terms, hence:  =(501⋅1+20012)i−(501⋅2+20022)−(500⋅3+19992)i+(500⋅4+20002)  =(501⋅1001)i−(501⋅1002)−(500⋅1001)i+(500⋅1002)  =((501−500)⋅1001)i−((501−500)⋅1002)  1001i−1002

Joseph Lewis · Accepted Answer

i+2i2+3i3+4i4+5i5+…+2002i2002=−1002+1001i  Explanation:  i+2i2+3i3+4i4+5i5+…+2002i2002=  =i−2−3i+4+5i−…−2002  =−2+4−6+8+…−2002+i(1−3+5−7+9−…+2001)  Now, grouping -2 with 4, and -6 with 8 and so forth, and similarly 1 with -3 and 5 with -7 and so on, that gives us  2+2+…+2−2002+i(−2−2−2−…−2+2001)  =−2002+2⋅500+i(2001−2⋅500)  =−1002+1001i

What is the sum of this imaginary number series problem: i+2i^{2}+3i^{3}+4i^{4}+5i^{5}+...+2002i^{2002}?

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