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

m
(solution (revised from paladin8's))
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
Let <math>\displaystyle f(n)</math> be the integer closest to <math>\displaystyle \sqrt[4]{n}.</math>  Find <math>\displaystyle \sum_{k=1}^{1995}\frac 1{f(k)}.</math>
+
Let <math>f(n)</math> be the integer closest to <math>\sqrt[4]{n}.</math>  Find <math>\sum_{k=1}^{1995}\frac 1{f(k)}.</math>
  
 
== Solution ==
 
== Solution ==
 +
When <math>\left(k - \frac {1}{2}\right)^4 < n < \left(k + \frac {1}{2}\right)^4</math>, then <math>f(n) = k</math>. Thus there are <math>\left \lfloor \left(k + \frac {1}{2}\right)^4 - \left(k - \frac {1}{2}\right)^4 \right\rfloor</math> values of <math>n</math> for which <math>f(n) = k</math>. Expanding using the [[binomial theorem]],
 +
 +
<cmath>\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*}</cmath>
 +
 +
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]]). 
 +
 +
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>370 + 30 = \boxed{400}</math>.
  
 
== See also ==
 
== See also ==
* [[1995_AIME_Problems/Problem_12|Previous Problem]]
+
{{AIME box|year=1995|num-b=12|num-a=14}}
* [[1995_AIME_Problems/Problem_14|Next Problem]]
+
 
* [[1995 AIME Problems]]
+
[[Category:Intermediate Algebra Problems]]

Revision as of 21:40, 29 July 2008

Problem

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

Solution

When $\left(k - \frac {1}{2}\right)^4 < n < \left(k + \frac {1}{2}\right)^4$, 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 $370 + 30 = \boxed{400}$.

See also

1995 AIME (ProblemsAnswer KeyResources)
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