2009 AMC 12A Problems/Problem 15

Problem

For what value of $n$ is $i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i$ ?

Note: here $i = \sqrt { - 1}$ .

$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 49 \qquad \textbf{(D)}\ 97 \qquad \textbf{(E)}\ 98$

Solution 1

We know that $i^x$ cycles every $4$ powers so we group the sum in $4$ s. $i+2i^2+3i^3+4i^4=2-2i$ $5i^5+6i^6+7i^7+8i^8=2-2i$

We can postulate that every group of $4$ is equal to $2-2i$ . For 24 groups we thus, get $48-48i$ as our sum. We know the solution must lie near The next term is the $24*4+1=97$ th term. This term is equal to $97i$ (first in a group of $4$ so $i^{97}=i$ ) and our sum is now $48+49i$ so $n=97\Rightarrow\boxed{\mathbf{D}}$ is our answer

Solution 2

Obviously, even powers of $i$ are real and odd powers of $i$ are imaginary. Hence the real part of the sum is $2i^2 + 4i^4 + 6i^6 + \ldots$ , and the imaginary part is $i + 3i^3 + 5i^5 + \cdots$ .

Let's take a look at the real part first. We have $i^2=-1$ , hence the real part simplifies to $-2+4-6+8-10+\cdots$ . If there were an odd number of terms, we could pair them as follows: $-2 + (4-6) + (8-10) + \cdots$ , hence the result would be negative. As we need the real part to be $48$ , we must have an even number of terms. If we have an even number of terms, we can pair them as $(-2+4) + (-6+8) + \cdots$ . Each parenthesis is equal to $2$ , thus there are $24$ of them, and the last value used is $96$ . This happens for $n=96$ and $n=97$ . As $n=96$ is not present as an option, we may conclude that the answer is $\boxed{97}$ .

In a complete solution, we should now verify which of $n=96$ and $n=97$ will give us the correct imaginary part.

We can rewrite the imaginary part as follows: $i + 3i^3 + 5i^5 + \cdots = i(1 + 3i^2 + 5i^4 + \cdots) = i(1 - 3 + 5 - \cdots)$ . We need to obtain $(1 - 3 + 5 - \cdots) = 49$ . Once again we can repeat the same reasoning: If the number of terms were even, the left hand side would be negative, thus the number of terms is odd. The left hand side can then be rewritten as $1 + (-3+5) + (-7+9) + \cdots$ . We need $24$ parentheses, therefore the last value used is $97$ . This happens when $n=97$ or $n=98$ , and we are done.

Solution 3 (Fast)

Some may know the equation:

$\sum_{k=1}^{n}kr^{k-1}=\frac{1-(n+1)r^n+nr^{n+1}}{(1-r)^2}$

(For those curious, this comes from differentiating the equation for finite geometric sums)

Using this equation, we have

$48+49i=i\frac{1-(n+1)i^n+ni^{n+1}}{(1-i)^2}$ $=\frac{1-(n+1)i^n+ni^{n+1}}{-2}$ $=-\frac{1}{2}+\frac{(n+1)i^n}{2}-\frac{ni^{n+1}}{2}$

Since the imaginary and the real part must be positive, we know that $i^{n+1}=-1$ or $i^{n+1}=-i$ . By the same line of reason, $i^{n}=1,i$ . This only works for $n\equiv 1 (\mod 4)$ . Therefore, we have:

$\frac{-1+n}{2}+\frac{(n+1)i}{2}=48+49i$

Solving either the real or imaginary part gives $\boxed{\mathbf{(D)}97}$

Video Solution

https://youtu.be/VfgUhcw112s

~savannahsolver

2009 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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