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Difference between revisions of "1995 AIME Problems/Problem 13"

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Thus, <math>\frac{1}{k}</math> appears in the summation <math>4k^3 + k</math> times, and the sum for each <math>k</math> is then <math>(4k^3 + k) \cdot \frac{1}{k} = 4k^2 + 1</math>. From <math>k = 1</math> to <math>k = 6</math>, we get <math>\sum_{k=1}^{6} 4k^2 + 1 = 364 + 6 = 370</math> (either adding or using the [[perfect square|sum of consecutive squares formula]]).   
 
Thus, <math>\frac{1}{k}</math> appears in the summation <math>4k^3 + k</math> times, and the sum for each <math>k</math> is then <math>(4k^3 + k) \cdot \frac{1}{k} = 4k^2 + 1</math>. From <math>k = 1</math> to <math>k = 6</math>, we get <math>\sum_{k=1}^{6} 4k^2 + 1 = 364 + 6 = 370</math> (either adding or using the [[perfect square|sum of consecutive squares formula]]).   
  
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But this only accounts for <math>\sum_{k = 1}^{6} (4k^3 + k) = 4\left(\frac{6(6+1)}{2}\right)^2 + \frac{6(6+1)}{2} = 1764 + 21 = 1785</math> terms, so we still have <math>1995 - 1785 = 210</math> terms with <math>f(n) = 7</math>. This adds <math>210 \cdot \frac {1}{7} = 30</math> to our summation, giving <math>{400}</math>.
+
But this only accounts for <math>\sum_{k = 1}^{6} (4k^3 + k) = 4\left(\frac{6(6+1)}{2}\right)^2 + \frac{6(6+1)}{2} = 1764 + 21 = 1785</math> terms, so we still have <math>1995 - 1785 = 210</math> terms with <math>f(n) = 7</math>. This adds <math>210 \cdot \frac {1}{7} = 30</math> to our summation, giving <math>\boxed{400}</math>.
  
 
==Solution 2==
 
==Solution 2==

Latest revision as of 23:07, 14 October 2025

Problem

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

Solution

When $\left(k - \frac {1}{2}\right)^4 \leq n < \left(k + \frac {1}{2}\right)^4$, $f(n) = k$. 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 $\boxed{400}$.

Solution 2

This is a pretty easy problem just to bash. Since the max number we can get is $7$, we just need to test $n$ values for $1.5,2.5,3.5,4.5,5.5$ and $6.5$. Then just do how many numbers there are times $\frac{1}{\lfloor n \rfloor}$, which should be $5+17+37+65+101+145+30 = \boxed{400}$

See also

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
All AIME Problems and Solutions

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