square

academic/misc/2024-math-calendar.html

@@ -181,7 +181,7 @@ <h1>Math Calendar 2024</h1>
         <td><a href="#mar18">18</a></td>
         <td><a href="#mar19">19</a></td>
         <td><a href="#mar20">20</a></td>
-        <td>21</td>
+        <td><a href="#mar21">21</a></td>
         <td>22</td>
         <td>23</td>
         <td>24</td>
@@ -3328,6 +3328,36 @@ <h3 id="mar20">Mar 20</h3>
     next to the 1, there are \(10\times2=20\) possible such labelings.
 </p>
 
+<h3 id="mar21">Mar 21</h3>
+
+<p>
+    The sum of the squares of five consecutive positive integers starting with
+    \(x\) is \(2655\).
+</p>
+
+<p>
+    We can solve this with the sum of squares formula which is:
+</p>
+
+\[\sum_{k=1}^nk^2={n(n+1)(2n+1)\over6}\]
+
+<p>
+    Let one summation have \(n=x+4\) and the other have \(n=x-1\). By
+    subtracting them, we get the sum \(x^2+(x+1)^2+(x+2)^2+(x+3)^2+(x+4)^2\).
+</p>
+
+\[\begin{align}&
+\sum_{k=1}^{x+4}k^2-\sum_{k=1}^{x-1}k^2
+={(x+4)(x+5)(2x+9)\over6}-{(x-1)x(2x-1)\over6}\\&
+={1\over6}(30x^2+120x+180)=5x^2+20x+30=2655
+\end{align}\]
+
+<p>
+    This gives us a quadratic equation: \(5x^2+20x-2625=5(x+25)(x-21)=0\). So
+    we must have \(x=21\), the positive solution. The other solution \(x=-25\)
+    works if we allow \(x\) to be negative.
+</p>
+
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