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

kihanja20

kihanja20

Answered

2022-01-18

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

Answer & Explanation

Matthew Rodriguez

Matthew Rodriguez

Expert

2022-01-19Added 32 answers

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:
=(5011+20012)i(5012+20022)(5003+19992)i+(5004+20002)
=(5011001)i(5011002)(5001001)i+(5001002)
=((501500)1001)i((501500)1002)
1001i1002
Joseph Lewis

Joseph Lewis

Expert

2022-01-20Added 43 answers

i+2i2+3i3+4i4+5i5++2002i2002=1002+1001i
Explanation:
i+2i2+3i3+4i4+5i5++2002i2002=
=i23i+4+5i2002
=2+46+8+2002+i(13+57+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++22002+i(2222+2001)
=2002+2500+i(20012500)
=1002+1001i

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?