1995 AIME Problems/Problem 13

Problem

Let $f(n)$ be the integer closest to $\sqrt[4]{n}.$ Find $\sum_{k=1}^{1995}\frac 1{f(k)}.$

Solution

When $\left(k - \frac {1}{2}\right)^4 \leq n < \left(k + \frac {1}{2}\right)^4$ , then $f(n) = k$ . Thus there are $\left \lfloor \left(k + \frac {1}{2}\right)^4 - \left(k - \frac {1}{2}\right)^4 \right\rfloor$ values of $n$ for which $f(n) = k$ . Expanding using the binomial theorem,

$\begin{align*} \left(k + \frac {1}{2}\right)^4 - \left(k - \frac {1}{2}\right)^4 &= \left(k^4 + 2k^3 + \frac 32k^2 + \frac 12k + \frac 1{16}\right) - \left(k^4 - 2k^3 + \frac 32k^2 - \frac 12k + \frac 1{16}\right)\\ &= 4k^3 + k. \end{align*}$

Thus, $\frac{1}{k}$ appears in the summation $4k^3 + k$ times, and the sum for each $k$ is then $(4k^3 + k) \cdot \frac{1}{k} = 4k^2 + 1$ . From $k = 1$ to $k = 6$ , we get $\sum_{k=1}^{6} 4k^2 + 1 = 364 + 6 = 370$ (either adding or using the sum of consecutive squares formula).

But this only accounts for $\sum_{k = 1}^{6} (4k^3 + k) = 4\left(\frac{6(6+1)}{2}\right)^2 + \frac{6(6+1)}{2} = 1764 + 21 = 1785$ terms, so we still have $1995 - 1785 = 210$ terms with $f(n) = 7$ . This adds $210 \cdot \frac {1}{7} = 30$ to our summation, giving ${400}$ .

1995 AIME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1995 AIME Problems/Problem 13

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