For what value of n is &i + 2i^2 + 3i^3 + … + ni^n &= 48 + 49i? Note: here i = √-1.

2009 AMC 12A Problem 15

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15.

For what value of

n

\begin{aligned} &i + 2i^2 + 3i^3 + \cdots + ni^n \\ &= 48 + 49i? \end{aligned}

Note: here

i = \sqrt{-1}.

24

48

49

97

98

Group the terms in blocks of four consecutive powers of

i

Each block

(k + 1)i^{k+1} + \cdots + (k + 4)i^{k+4}

with

k

a multiple of

4

equals

2 - 2i

For

k

a multiple of

4,

\begin{aligned} &(k + 1)i^{k+1} + (k + 2)i^{k+2} \\ &\quad {}+ (k + 3)i^{k+3} + (k + 4)i^{k+4} \\ &= (k + 1)i - (k + 2) \\ &\quad {}- (k + 3)i + (k + 4) \\ &= 2 - 2i. \end{aligned}

Summing the first

96

terms (that is

24

blocks) gives

24(2 - 2i) = 48 - 48i.

Adding the next term

97i^{97} = 97i

yields

48 - 48i + 97i = 48 + 49i.

n = 97.

Thus, the correct answer is D.