Algowikiadmin: Recovering wiki

2014-09-11T18:22:56Z

Recovering wiki

Algowikiadmin: Recovering wiki

2014-09-11T18:13:37Z

Recovering wiki

New page

Step 1: Show that the statement holds for the basis case <math>n = 1</math> 
:<math>\frac {1}{i(i+1)} = \frac {n}{n+1}</math> 

:<math>\frac {1}{1(1+1)} = \frac {1}{1+1}</math> 

:<math>\frac {1}{2} = \frac {1}{2}</math> 

Since <math>1/2 = 1/2</math>, the basis case is true. 

Step 2: Assume that that summation is true up to ''n''. 
Step 3: Show that on the assumption that the summation is true for ''n'', it follows that it is true for ''n + 1''.

<math>\sum_{i = 1}^{n+1} = \frac{n+1}{n+1+1} = \frac{n}{n+1} + \frac{1}{(n+1)(n+1+1)}</math> 
<math>\frac{n+1}{n+2} = \frac{n(n+2)}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)}</math> 
<math>\frac{n+1}{n+2} = \frac{n(n+2)+1}{(n+1)(n+2)}</math> 
<math>\frac{n+1}{n+2} = \frac{n^2+2n+1}{(n+1)(n+2)}</math> 
<math>\frac{n+1}{n+2} = \frac{(n+1)(n+1)}{(n+1)(n+2)}</math> 
<math>\frac{n+1}{n+2} = \frac{(n+1)}{(n+2)}</math> 

QED

@@ Line 1: / Line 1: @@
-&lt;b&gt;Step 1:&lt;/b&gt; Show that the statement holds for the basis case &lt;math&gt;n = 1&lt;/math&gt;&lt;br&gt;
+<b>Step 1:</b> Show that the statement holds for the basis case <math>n = 1</math><br>
-:&lt;math&gt;\frac {1}{i(i+1)} = \frac {n}{n+1}&lt;/math&gt;&lt;br&gt;&lt;br&gt;
+:<math>\frac {1}{i(i+1)} = \frac {n}{n+1}</math><br><br>
-:&lt;math&gt;\frac {1}{1(1+1)} = \frac {1}{1+1}&lt;/math&gt;&lt;br&gt;&lt;br&gt;
+:<math>\frac {1}{1(1+1)} = \frac {1}{1+1}</math><br><br>
-:&lt;math&gt;\frac {1}{2} = \frac {1}{2}&lt;/math&gt;&lt;br&gt;&lt;br&gt;
+:<math>\frac {1}{2} = \frac {1}{2}</math><br><br>
-Since &lt;math&gt;1/2 = 1/2&lt;/math&gt;, the basis case is true.&lt;br&gt;&lt;br&gt;
+Since <math>1/2 = 1/2</math>, the basis case is true.<br><br>
-&lt;b&gt;Step 2:&lt;/b&gt; Assume that that summation is true up to ''n''.&lt;br&gt;&lt;br&gt;
+<b>Step 2:</b> Assume that that summation is true up to ''n''.<br><br>
-&lt;b&gt;Step 3:&lt;/b&gt; Show that on the assumption that the summation is true for ''n'', it follows that it is true for ''n + 1''.
+<b>Step 3:</b> Show that on the assumption that the summation is true for ''n'', it follows that it is true for ''n + 1''.
-&lt;math&gt;\sum_{i = 1}^{n+1} = \frac{n+1}{n+1+1} = \frac{n}{n+1} + \frac{1}{(n+1)(n+1+1)}&lt;/math&gt;&lt;br&gt;
+<math>\sum_{i = 1}^{n+1} = \frac{n+1}{n+1+1} = \frac{n}{n+1} + \frac{1}{(n+1)(n+1+1)}</math><br>
-&lt;math&gt;\frac{n+1}{n+2} = \frac{n(n+2)}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)}&lt;/math&gt;&lt;br&gt;
+<math>\frac{n+1}{n+2} = \frac{n(n+2)}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)}</math><br>
-&lt;math&gt;\frac{n+1}{n+2} = \frac{n(n+2)+1}{(n+1)(n+2)}&lt;/math&gt;&lt;br&gt;
+<math>\frac{n+1}{n+2} = \frac{n(n+2)+1}{(n+1)(n+2)}</math><br>
-&lt;math&gt;\frac{n+1}{n+2} = \frac{n^2+2n+1}{(n+1)(n+2)}&lt;/math&gt;&lt;br&gt;
+<math>\frac{n+1}{n+2} = \frac{n^2+2n+1}{(n+1)(n+2)}</math><br>
-&lt;math&gt;\frac{n+1}{n+2} = \frac{(n+1)(n+1)}{(n+1)(n+2)}&lt;/math&gt;&lt;br&gt;
+<math>\frac{n+1}{n+2} = \frac{(n+1)(n+1)}{(n+1)(n+2)}</math><br>
-&lt;math&gt;\frac{n+1}{n+2} = \frac{(n+1)}{(n+2)}&lt;/math&gt;&lt;br&gt;
+<math>\frac{n+1}{n+2} = \frac{(n+1)}{(n+2)}</math><br>
 QED

TADM2E 1.15 - Revision history

Algowikiadmin: Recovering wiki

Algowikiadmin: Recovering wiki