<?xml version="1.0" encoding="utf-8"?>
<?xml-stylesheet type="text/xsl" href="../../assets/xml/rss.xsl" media="all"?><rss version="2.0" xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>Calculemus (Publicaciones sobre Summation)</title><link>https://jaalonso.github.io/calculemus/</link><description></description><atom:link href="https://jaalonso.github.io/calculemus/es/categories/cat_summation.xml" rel="self" type="application/rss+xml"></atom:link><language>es</language><copyright>Contents © 2026 &lt;a href="mailto:"&gt;José A. Alonso&lt;/a&gt; 
&lt;a rel="license" href="https://creativecommons.org/licenses/by-nc-sa/4.0/"&gt;
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style="border-width:0; margin-bottom:12px;"
src="https://i.creativecommons.org/l/by-nc-sa/4.0/88x31.png"&gt;&lt;/a&gt;</copyright><lastBuildDate>Wed, 14 Oct 2026 18:49:42 GMT</lastBuildDate><generator>Nikola (getnikola.com)</generator><docs>http://blogs.law.harvard.edu/tech/rss</docs><item><title>Proofs of ∑k&lt;n. 2ᵏ = 2ⁿ-1</title><link>https://jaalonso.github.io/calculemus/es/posts/2024/09/24-sum_of_powers_of_two/</link><dc:creator>José A. Alonso</dc:creator><description>&lt;div&gt;&lt;p&gt;Prove that
\[ 1 + 2 + 2^2 + 2^3 + ... + 2^{n-1} = 2^n - 1 \]&lt;/p&gt;
&lt;p&gt;To do this, complete the following Lean4 theory:&lt;/p&gt;
&lt;div class="code"&gt;&lt;pre class="code literal-block"&gt;&lt;span class="kn"&gt;import&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="n"&gt;Mathlib.Tactic&lt;/span&gt;

&lt;span class="kn"&gt;open&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="n"&gt;Finset&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="n"&gt;Nat&lt;/span&gt;

&lt;span class="kd"&gt;variable&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="o"&gt;(&lt;/span&gt;&lt;span class="n"&gt;n&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="o"&gt;:&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="n"&gt;ℕ&lt;/span&gt;&lt;span class="o"&gt;)&lt;/span&gt;

&lt;span class="kd"&gt;example&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="o"&gt;:&lt;/span&gt;
&lt;span class="w"&gt;  &lt;/span&gt;&lt;span class="bp"&gt;∑&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="n"&gt;k&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="k"&gt;in&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="n"&gt;range&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="n"&gt;n&lt;/span&gt;&lt;span class="o"&gt;,&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="mi"&gt;2&lt;/span&gt;&lt;span class="bp"&gt;^&lt;/span&gt;&lt;span class="n"&gt;k&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="bp"&gt;=&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="mi"&gt;2&lt;/span&gt;&lt;span class="bp"&gt;^&lt;/span&gt;&lt;span class="n"&gt;n&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="bp"&gt;-&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="mi"&gt;1&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="o"&gt;:=&lt;/span&gt;
&lt;span class="kd"&gt;by&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="gr"&gt;sorry&lt;/span&gt;
&lt;/pre&gt;&lt;/div&gt;

&lt;p&gt;&lt;a href="https://jaalonso.github.io/calculemus/es/posts/2024/09/24-sum_of_powers_of_two/"&gt;Leer más…&lt;/a&gt; (quedan 1 minutos de lectura)&lt;/p&gt;&lt;/div&gt;</description><guid>https://jaalonso.github.io/calculemus/es/posts/2024/09/24-sum_of_powers_of_two/</guid><pubDate>Tue, 24 Sep 2024 04:00:00 GMT</pubDate></item><item><title>Proofs of "∑i&lt;n. i = n(n-1)/2"</title><link>https://jaalonso.github.io/calculemus/es/posts/2024/09/19-gausss_formula_for_summation/</link><dc:creator>José A. Alonso</dc:creator><description>&lt;div&gt;&lt;p&gt;Gauss's formula for the sum of the first natural numbers is
\[ 0 + 1 + 2 + ... + (n-1) = \dfrac{n(n-1)}{2} \]
that is,
\[ \sum_{i &amp;lt; n} i = \dfrac{n(n-1)}{2} \]&lt;/p&gt;
&lt;p&gt;In a previous exercise, this formula was proven by induction. Another way to prove it, without using induction, is the following: The sum can be written in two ways
\begin{align}
   S &amp;amp;= &amp;amp;0     &amp;amp;+ &amp;amp;1     &amp;amp;+ &amp;amp;2     &amp;amp;+ ... &amp;amp;+ &amp;amp;(n-3) &amp;amp;+ &amp;amp;(n-2) &amp;amp;+ &amp;amp;(n-1) \newline
   S &amp;amp;= &amp;amp;(n-1) &amp;amp;+ &amp;amp;(n-2) &amp;amp;+ &amp;amp;(n-3) &amp;amp;+ ... &amp;amp;+ &amp;amp;2     &amp;amp;+ &amp;amp;1     &amp;amp;+ &amp;amp;0
\end{align}
By adding, we observe that each pair of numbers in the same column sums to (n-1), and since there are n columns in total, it follows that
\[ 2S = n(n-1) \]
which proves the formula.&lt;/p&gt;
&lt;p&gt;Prove Gauss's formula by following the above procedure.&lt;/p&gt;
&lt;p&gt;To do this, complete the following Lean4 theory:&lt;/p&gt;
&lt;div class="code"&gt;&lt;pre class="code literal-block"&gt;&lt;span class="kn"&gt;import&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="n"&gt;Mathlib.Algebra.BigOperators.Intervals&lt;/span&gt;
&lt;span class="kn"&gt;import&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="n"&gt;Mathlib.Tactic&lt;/span&gt;

&lt;span class="kn"&gt;open&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="n"&gt;Finset&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="n"&gt;Nat&lt;/span&gt;

&lt;span class="kd"&gt;variable&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="o"&gt;(&lt;/span&gt;&lt;span class="n"&gt;n&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="n"&gt;i&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="o"&gt;:&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="n"&gt;ℕ&lt;/span&gt;&lt;span class="o"&gt;)&lt;/span&gt;

&lt;span class="kd"&gt;example&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="o"&gt;:&lt;/span&gt;
&lt;span class="w"&gt;  &lt;/span&gt;&lt;span class="bp"&gt;∑&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="n"&gt;i&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="bp"&gt;∈&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="n"&gt;range&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="n"&gt;n&lt;/span&gt;&lt;span class="o"&gt;,&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="n"&gt;i&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="bp"&gt;=&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="n"&gt;n&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="bp"&gt;*&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="o"&gt;(&lt;/span&gt;&lt;span class="n"&gt;n&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="bp"&gt;-&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="mi"&gt;1&lt;/span&gt;&lt;span class="o"&gt;)&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="bp"&gt;/&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="mi"&gt;2&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="o"&gt;:=&lt;/span&gt;
&lt;span class="kd"&gt;by&lt;/span&gt;&lt;span class="w"&gt; &lt;/span&gt;&lt;span class="gr"&gt;sorry&lt;/span&gt;
&lt;/pre&gt;&lt;/div&gt;

&lt;p&gt;&lt;a href="https://jaalonso.github.io/calculemus/es/posts/2024/09/19-gausss_formula_for_summation/"&gt;Leer más…&lt;/a&gt; (quedan 1 minutos de lectura)&lt;/p&gt;&lt;/div&gt;</description><guid>https://jaalonso.github.io/calculemus/es/posts/2024/09/19-gausss_formula_for_summation/</guid><pubDate>Thu, 19 Sep 2024 04:00:00 GMT</pubDate></item></channel></rss>