2009 AMC 12A Problem 15

Attempt Problem 15 of the 2009 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 12A solutions, or check the answer key.

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15.

For what value of $n$ is $\begin{aligned} &i + 2i^2 + 3i^3 + \cdots + ni^n \\ &= 48 + 49i? \end{aligned}$ Note: here $i = \sqrt{-1}.$

$24$

$48$

$49$

$97$

$98$

Answer: D

Concepts:complex number pairing and grouping

Difficulty rating: 2010

Solution:

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 $48 - 48i + 97i = 48 + 49i.$ So $n = 97.$

Thus, the correct answer is D.

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