Question

If $n$ is a multiple of $4,$ the sum

$\begin{aligned} s&=1+2i+3i^2+\cdots\\ &\quad+(n+1)i^n, \end{aligned}$

where $i=\sqrt{-1},$ equals:

$1+i$

$\dfrac12(n+2)$

$\dfrac12(n+2-ni)$

$\dfrac12\left[(n+1)(1-i)+2\right]$

$\dfrac18(n^2+8-4ni)$

Answer: C

Concepts:complex number roots of unity summation

Difficulty rating: 1870

Small Hint:

Group the terms in blocks of four powers of $i$

Big Hint:

Each complete block beginning with coefficient $4j+1$ sums to $-2-2i$

Solution:

Write $n=4m.$ For $j=0,\ldots,m-1,$ the four terms with exponents $4j$ through $4j+3$ sum to $\begin{aligned} &(4j+1)+(4j+2)i\\ &\quad-(4j+3)-(4j+4)i\\ &=-2-2i. \end{aligned}$ The final term is $4m+1.$ Therefore $\begin{aligned} s&=m(-2-2i)+(4m+1)\\ &=\frac{n+2-ni}{2}. \end{aligned}$

Thus, the correct answer is C.

1964 AMC 12 Problem 34

Answer: C

Difficulty rating: 1870

Small Hint:

Big Hint:

Solution:

Problem 34 in Other Years