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2009 AMC 12A Problem 15

Problem 15 of 25IntermediateAlgebraProblem-Solving Techniques

For what value of nn is i+2i2+3i3+⋯+nin=48+49i? \begin{aligned} &i + 2i^2 + 3i^3 + \cdots + ni^n \\ &= 48 + 49i? \end{aligned} Note: here i=−1.i = \sqrt{-1}.

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Solution

For kk a multiple of 4,4, (k+1)ik+1+(k+2)ik+2+(k+3)ik+3+(k+4)ik+4=(k+1)i−(k+2)−(k+3)i+(k+4)=2−2i. \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 9696 terms (that is 2424 blocks) gives 24(2−2i)=48−48i.24(2 - 2i) = 48 - 48i. Adding the next term 97i97=97i97i^{97} = 97i yields 48−48i+97i=48+49i.48 - 48i + 97i = 48 + 49i. So n=97.n = 97. Thus, the correct answer is D.
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