kihanja20

2022-01-18

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

Matthew Rodriguez

Expert

The Complex multipliers can be simplified into a cyclic pattern of period 4:
$i=i$
${i}^{2}=-1$
${i}^{3}=-i$
${i}^{4}=1$
Since ${i}^{4}=1$, we have etc. and we find:
$\sum _{n=1}^{2002}n{i}^{n}$
$=\left(\sum _{k=0}^{500}\left(4k+1\right)\right)i-\left(\sum _{k=0}^{500}\left(4k+2\right)\right)-\left(\sum _{k=0}^{499}\left(4k+3\right)\right)i+\left(\sum _{k=0}^{499}\left(4k+4\right)\right)$
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:
$=\left(501\cdot \frac{1+2001}{2}\right)i-\left(501\cdot \frac{2+2002}{2}\right)-\left(500\cdot \frac{3+1999}{2}\right)i+\left(500\cdot \frac{4+2000}{2}\right)$
$=\left(501\cdot 1001\right)i-\left(501\cdot 1002\right)-\left(500\cdot 1001\right)i+\left(500\cdot 1002\right)$
$=\left(\left(501-500\right)\cdot 1001\right)i-\left(\left(501-500\right)\cdot 1002\right)$
$1001i-1002$

Joseph Lewis

Expert

$i+2{i}^{2}+3{i}^{3}+4{i}^{4}+5{i}^{5}+\dots +2002{i}^{2002}=-1002+1001i$
Explanation:
$i+2{i}^{2}+3{i}^{3}+4{i}^{4}+5{i}^{5}+\dots +2002{i}^{2002}=$
$=i-2-3i+4+5i-\dots -2002$
$=-2+4-6+8+\dots -2002+i\left(1-3+5-7+9-\dots +2001\right)$
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
$2+2+\dots +2-2002+i\left(-2-2-2-\dots -2+2001\right)$
$=-2002+2\cdot 500+i\left(2001-2\cdot 500\right)$
$=-1002+1001i$

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