Question

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

Answer 1

`1001i - 1002`

Explanation:

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 `i = i^5 = i^9 =...` and `i^2 = i^6 = i^10 =...`, etc. and we find:

`sum_(n=1)^2002 n i^n`

`=(sum_(k=0)^500 (4k+1)) i - (sum_(k=0)^500 (4k+2)) - (sum_(k=0)^499 (4k+3)) i + (sum_(k=0)^499 (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+2001)/2) i - (501 * (2+2002)/2) - (500 * (3+1999)/2) i + (500 * (4+2000)/2)`

`=(501*1001) i - (501*1002) - (500*1001) i + (500*1002)`

`=((501-500)*1001) i - ((501-500)*1002)`

`=1001i - 1002`

Answer 2

`i+2i^2+3i^3+4i^4+5i^5+...+2002i^2002=-1002+1001i`

Explanation:

`i+2i^2+3i^3+4i^4+5i^5+...+2002i^2002=`

`=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`