Problem:
For what value of n is i+2i2+3i3+⋯+nin=48+49i ? Note: here i=−1.
Answer Choices:
A. 24
B. 48
C. 49
D. 97
E. 98
Solution:
Let k be a multiple of 4 . For k≥0,
(k+1)ik+1+(k+2)ik+2+(k+3)ik+3+(k+4)ik+4=(k+1)i+(k+2)(−1)+(k+3)(−i)+(k+4)=2−2i
Thus when n=4⋅24=96, we have i+2i2+⋯+nin=24(2−2i)=48−48i. Adding the term 97i97=97i gives (48−48i)+97i=48+49i when n=97.
The problems on this page are the property of the MAA's American Mathematics Competitions