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Difference between revisions of "1968 IMO Problems/Problem 6"
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== Solution 3 ==
 
== Solution 3 ==
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By Hermite's identity, <cmath>\bigg[ n + \frac12 \bigg] = \bigg[ 2n \bigg] - \bigg[ n \bigg].</cmath>
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By Hermite's identity, for real numbers <math>x,</math> <cmath>\bigg[ x + \frac12 \bigg] = \bigg[ 2x \bigg] - \bigg[ x \bigg].</cmath>
  
−
Hence our sum telescopes: <cmath>\sum_{k = 0}^\infty\bigg[\frac{n + 2^k}{2^{k + 1}}\bigg] = \sum_{k = 0}^\infty\bigg[\frac{n}{2^{k + 1}} + \frac12 \bigg] = \sum_{k = 0}^\infty\bigg[\frac{n}{2^{k}} - \frac{n}{2^{k+1}} \bigg] = \bigg[ n \bigg].</cmath>
+
Hence our sum telescopes: <cmath>\sum_{k = 0}^\infty\bigg[\frac{n + 2^k}{2^{k + 1}}\bigg] = \sum_{k = 0}^\infty\bigg[\frac{n}{2^{k + 1}} + \frac12 \bigg] = \sum_{k = 0}^\infty \left( \bigg[\frac{n}{2^{k}} \bigg] - \bigg[ \frac{n}{2^{k+1}} \bigg] \right) = \bigg[ n \bigg].</cmath>
  
 
~Maximilian113
 
~Maximilian113

Latest revision as of 11:49, 5 December 2023

Problem

For every natural number $n$, evaluate the sum \[\sum_{k = 0}^\infty\bigg[\frac{n + 2^k}{2^{k + 1}}\bigg] = \Big[\frac{n + 1}{2}\Big] + \Big[\frac{n + 2}{4}\Big] + \cdots + \bigg[\frac{n + 2^k}{2^{k + 1}}\bigg] + \cdots\] (The symbol $[x]$ denotes the greatest integer not exceeding $x$.)

Solution

I shall prove that the summation is equal to $n$.

Let the binary representation of $n$ be $\overline{d_xd_{x-1}\dots d_1d_0}_2$, where $d_i\in\{0,1\}$ for all $i$, and $x=\lfloor \log_2 n\rfloor$. Note that if $d_i=0$, then $\Big[\frac{n + 2^i}{2^{i+1}}\Big]=\Big[\frac{n}{2^{i+1}}\Big]$; and if $d_i=1$, then $\Big[\frac{n + 2^i}{2^{i+1}}\Big]=\Big[\frac{n}{2^{i+1}}\Big]+1$. Also note that $\Big[\frac{n + 2^i}{2^{i+1}}\Big]=0$ for all $i\geq x+1$. Therefore the given sum is equal to

\[\sum_{k = 0}^\infty\bigg[\frac{n + 2^k}{2^{k + 1}}\bigg] =S+\sum_{k=0}^{x} \bigg[\frac{n}{2^{k + 1}}\bigg]\]

where $S$ is the number of 1's in the binary representation of $n$. Legendre's Formula states that $n-S=\sum_{k=0}^{x} \bigg[\frac{n}{2^{k + 1}}\bigg]$, which proves the assertion. $\blacksquare$


Solution 2

We observe \[n \equiv 2^k \text{mod }2^{k+1} \longrightarrow \left[\frac{n+1+2^k}{2^{k+1}}\right] - \left[\frac{n+2^k}{2^{k+1}}\right] = 1 \text{ and } \left[\frac{n+1+2^k}{2^{k+1}}\right] - \left[\frac{n+2^k}{2^{k+1}}\right] = 0 \text{ otherwise.}\]

But \[D = [d \text{ } | \text{ } \exists \text{ } s \in \mathbb{N} \text{ such that } d \equiv 2^{s-1} \text{ mod } 2^s] = \mathbb{N} \longrightarrow \sum_{k = 0}^\infty\bigg[\frac{n+1 + 2^k}{2^{k + 1}}\bigg] - \sum_{k = 0}^\infty\bigg[\frac{n + 2^k}{2^{k + 1}}\bigg] = 1,\]

so the result is just $n$. $\square$

~ilovepi3.14


Solution 3

By Hermite's identity, for real numbers $x,$ \[\bigg[ x + \frac12 \bigg] = \bigg[ 2x \bigg] - \bigg[ x \bigg].\]

Hence our sum telescopes: \[\sum_{k = 0}^\infty\bigg[\frac{n + 2^k}{2^{k + 1}}\bigg] = \sum_{k = 0}^\infty\bigg[\frac{n}{2^{k + 1}} + \frac12 \bigg] = \sum_{k = 0}^\infty \left( \bigg[\frac{n}{2^{k}} \bigg] - \bigg[ \frac{n}{2^{k+1}} \bigg] \right) = \bigg[ n \bigg].\]

~Maximilian113

Resources

1968 IMO (Problems) • Resources
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