TADM2E 1.13 - Revision history

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Call the statement <math>S_n</math> and the general term <math>a_n</math> 

Step 1: Show that the statement holds for the basis case <math>n = 0</math> 

:<math>a_0 = 0 \cdot (0 + 1)(0 + 2) = 0</math>
 
:<math>S_0 = \frac {0 \cdot (0 + 1)(0 + 2)(0 + 3)} {4} = \frac {0} {4} = 0 </math> 

Since <math>a_n = S_n</math>, the basis case is true. 

Step 2: Assume that <math>n = k</math> holds.
:<math>S_k = \frac {k(k + 1)(k + 2)(k + 3)} {4}</math> 

Step 3: Show that on the assumption that the summation is true for ''k'', it follows that it is true for ''k + 1''.

:<math>
\begin{align}
S_{k + 1} & = \sum_{i = 1}^k i(i + 1)(i + 2) + a_{k + 1} \\
& = \frac{k(k + 1)(k + 2)(k + 3)} {4} + (k + 1)((k + 1) + 1)((k + 1) + 2) \\
& = \frac{k(k + 1)(k + 2)(k + 3)} {4} + (k + 1)(k + 2)(k + 3) \\
& = \frac{k(k + 1)(k + 2)(k + 3)}{4} + \frac{4 \cdot (k + 1)(k + 2)(k + 3)} {4} \\
\end{align}
</math> 

It's easier to factor than expand. Notice the common factor of ''(k + 1)(k + 2)(k + 3)''.
:<math>S_{k + 1} = \frac{(k + 1)(k + 2)(k + 3)(k + 4)}{4}</math> 

This should be equal to the formula
<math>S_k = \frac{k(k + 1)(k + 2)(k + 3)}{4}</math>
when ''k = k + 1'': 
:<math>\frac{(k + 1)((k + 1) + 1)((k + 2) + 2)((k + 1) + 3)} {4} =
\frac{(k + 1)(k + 2)(k + 3)(k + 4)} {4}</math> 

Since both the basis and the inductive step have been proved, it has now been proved by mathematical induction that S_n holds for all natural n

[[introduction-TADM2E|Back to ''Introduction ...'']]

← Older revision		Revision as of 00:51, 1 August 2020
Line 1:		Line 1:
−	~~#REDIRECT~~ [[~~UNTV~~]]	+	Call the statement <math>S_n</math> and the general term <math>a_n</math><br>
		+
		+	<b>Step 1:</b> Show that the statement holds for the basis case <math>n = 0</math><br>
		+
		+	:<math>a_0 = 0 \cdot (0 + 1)(0 + 2) = 0</math>
		+	<br><br>
		+	:<math>S_0 = \frac {0 \cdot (0 + 1)(0 + 2)(0 + 3)} {4} = \frac {0} {4} = 0 </math><br>
		+
		+	Since <math>a_n = S_n</math>, the basis case is true.<br><br>
		+
		+	<b>Step 2:</b> Assume that <math>n = k</math> holds.
		+	:<math>S_k = \frac {k(k + 1)(k + 2)(k + 3)} {4}</math><br><br>
		+
		+	<b>Step 3:</b> Show that on the assumption that the summation is true for ''k'', it follows that it is true for ''k + 1''.
		+
		+	:<math>
		+	\begin{align}
		+	S_{k + 1} & = \sum_{i = 1}^k i(i + 1)(i + 2) + a_{k + 1} \\
		+	& = \frac{k(k + 1)(k + 2)(k + 3)} {4} + (k + 1)((k + 1) + 1)((k + 1) + 2) \\
		+	& = \frac{k(k + 1)(k + 2)(k + 3)} {4} + (k + 1)(k + 2)(k + 3) \\
		+	& = \frac{k(k + 1)(k + 2)(k + 3)}{4} + \frac{4 \cdot (k + 1)(k + 2)(k + 3)} {4} \\
		+	\end{align}
		+	</math><br>
		+
		+	It's easier to factor than expand. Notice the common factor of ''(k + 1)(k + 2)(k + 3)''.
		+	:<math>S_{k + 1} = \frac{(k + 1)(k + 2)(k + 3)(k + 4)}{4}</math><br>
		+
		+	This should be equal to the formula
		+	<math>S_k = \frac{k(k + 1)(k + 2)(k + 3)}{4}</math>
		+	when ''k = k + 1'':<br>
		+	:<math>\frac{(k + 1)((k + 1) + 1)((k + 2) + 2)((k + 1) + 3)} {4} =
		+	\frac{(k + 1)(k + 2)(k + 3)(k + 4)} {4}</math><br><br>
		+
		+	Since both the basis and the inductive step have been proved, it has now been proved by mathematical induction that S_n holds for all natural n
		+
		+	[[introduction-TADM2E\|Back to ''Introduction ...'']]

← Older revision		Revision as of 10:13, 31 July 2020
Line 1:		Line 1:
−	~~Call the statement <math>S_n</math> and the general term <math>a_n</math><br>~~	+	#REDIRECT [[UNTV]]
−
−	~~<b>Step 1:</b> Show that the statement holds for the basis case <math>n = 0</math><br>~~
−
−	~~:<math>a_0 = 0 \cdot (0 + 1)(0 + 2) = 0</math>~~
−	~~<br><br>~~
−	~~:<math>S_0 = \frac {0 \cdot (0 + 1)(0 + 2)(0 + 3)} {4} = \frac {0} {4} = 0 </math><br>~~
−
−	~~Since <math>a_n = S_n</math>, the basis case is true.<br><br>~~
−
−	~~<b>Step 2:</b> Assume that <math>n = k</math> holds.~~
−	~~:<math>S_k = \frac {k(k + 1)(k + 2)(k + 3)} {4}</math><br><br>~~
−
−	~~<b>Step 3:</b> Show that on the assumption that the summation is true for ''k'', it follows that it is true for ''k + 1''.~~
−
−	~~:<math>~~
−	~~\begin{align}~~
−	~~S_{k + 1} & = \sum_{i = 1}^k i(i + 1)(i + 2) + a_{k + 1} \\~~
−	~~& = \frac{k(k + 1)(k + 2)(k + 3)} {4} + (k + 1)((k + 1) + 1)((k + 1) + 2) \\~~
−	~~& = \frac{k(k + 1)(k + 2)(k + 3)} {4} + (k + 1)(k + 2)(k + 3) \\~~
−	~~& = \frac{k(k + 1)(k + 2)(k + 3)}{4} + \frac{4 \cdot (k + 1)(k + 2)(k + 3)} {4} \\~~
−	~~\end{align}~~
−	~~</math><br>~~
−
−	~~It's easier to factor than expand. Notice the common factor of ''(k + 1)(k + 2)(k + 3)''.~~
−	~~:<math>S_{k + 1} = \frac{(k + 1)(k + 2)(k + 3)(k + 4)}{4}</math><br>~~
−
−	~~This should be equal to the formula~~
−	~~<math>S_k = \frac{k(k + 1)(k + 2)(k + 3)}{4}</math>~~
−	~~when ''k = k + 1'':<br>~~
−	~~:<math>\frac{(k + 1)((k + 1) + 1)((k + 2) + 2)((k + 1) + 3)} {4} =~~
−	~~\frac{(k + 1)(k + 2)(k + 3)(k + 4)} {4}</math><br><br>~~
−
−	~~Since both the basis and the inductive step have been proved, it has now been proved by mathematical induction that S_n holds for all natural n~~
−
−	[[~~introduction-TADM2E\|Back to ''Introduction ...''~~]]

@@ Line 1: / Line 1: @@
-Call the statement &lt;math&gt;S_n&lt;/math&gt; and the general term &lt;math&gt;a_n&lt;/math&gt;&lt;br&gt;
+Call the statement <math>S_n</math> and the general term <math>a_n</math><br>
-&lt;b&gt;Step 1:&lt;/b&gt; Show that the statement holds for the basis case &lt;math&gt;n = 0&lt;/math&gt;&lt;br&gt;
+<b>Step 1:</b> Show that the statement holds for the basis case <math>n = 0</math><br>
-:&lt;math&gt;a_0 = 0 \cdot (0 + 1)(0 + 2) = 0&lt;/math&gt;
+:<math>a_0 = 0 \cdot (0 + 1)(0 + 2) = 0</math>
-&lt;br&gt;&lt;br&gt;
+<br><br>
-:&lt;math&gt;S_0 = \frac {0 \cdot (0 + 1)(0 + 2)(0 + 3)} {4} = \frac {0} {4} = 0 &lt;/math&gt;&lt;br&gt;
+:<math>S_0 = \frac {0 \cdot (0 + 1)(0 + 2)(0 + 3)} {4} = \frac {0} {4} = 0 </math><br>
-Since &lt;math&gt;a_n = S_n&lt;/math&gt;, the basis case is true.&lt;br&gt;&lt;br&gt;
+Since <math>a_n = S_n</math>, the basis case is true.<br><br>
-&lt;b&gt;Step 2:&lt;/b&gt; Assume that &lt;math&gt;n = k&lt;/math&gt; holds.
+<b>Step 2:</b> Assume that <math>n = k</math> holds.
-:&lt;math&gt;S_k = \frac {k(k + 1)(k + 2)(k + 3)} {4}&lt;/math&gt;&lt;br&gt;&lt;br&gt;
+:<math>S_k = \frac {k(k + 1)(k + 2)(k + 3)} {4}</math><br><br>
-&lt;b&gt;Step 3:&lt;/b&gt; Show that on the assumption that the summation is true for ''k'', it follows that it is true for ''k + 1''.
+<b>Step 3:</b> Show that on the assumption that the summation is true for ''k'', it follows that it is true for ''k + 1''.
-:&lt;math&gt;
+:<math>
 \begin{align}
-S_{k + 1} &amp; = \sum_{i = 1}^k i(i + 1)(i + 2) + a_{k + 1} \\
+S_{k + 1} & = \sum_{i = 1}^k i(i + 1)(i + 2) + a_{k + 1} \\
-           &amp; = \frac{k(k + 1)(k + 2)(k + 3)} {4} + (k + 1)((k + 1) + 1)((k + 1) + 2) \\
+           & = \frac{k(k + 1)(k + 2)(k + 3)} {4} + (k + 1)((k + 1) + 1)((k + 1) + 2) \\
-           &amp; = \frac{k(k + 1)(k + 2)(k + 3)} {4} + (k + 1)(k + 2)(k + 3) \\
+           & = \frac{k(k + 1)(k + 2)(k + 3)} {4} + (k + 1)(k + 2)(k + 3) \\
-           &amp; = \frac{k(k + 1)(k + 2)(k + 3)}{4} + \frac{4 \cdot (k + 1)(k + 2)(k + 3)} {4} \\
+           & = \frac{k(k + 1)(k + 2)(k + 3)}{4} + \frac{4 \cdot (k + 1)(k + 2)(k + 3)} {4} \\
 \end{align}
-&lt;/math&gt;&lt;br&gt;
+</math><br>
 It's easier to factor than expand. Notice the common factor of ''(k + 1)(k + 2)(k + 3)''.
-:&lt;math&gt;S_{k + 1} = \frac{(k + 1)(k + 2)(k + 3)(k + 4)}{4}&lt;/math&gt;&lt;br&gt;
+:<math>S_{k + 1} = \frac{(k + 1)(k + 2)(k + 3)(k + 4)}{4}</math><br>
 This should be equal to the formula
-&lt;math&gt;S_k = \frac{k(k + 1)(k + 2)(k + 3)}{4}&lt;/math&gt;
+<math>S_k = \frac{k(k + 1)(k + 2)(k + 3)}{4}</math>
-when ''k = k + 1'':&lt;br&gt;
+when ''k = k + 1'':<br>
-:&lt;math&gt;\frac{(k + 1)((k + 1) + 1)((k + 2) + 2)((k + 1) + 3)} {4} =
+:<math>\frac{(k + 1)((k + 1) + 1)((k + 2) + 2)((k + 1) + 3)} {4} =
-\frac{(k + 1)(k + 2)(k + 3)(k + 4)} {4}&lt;/math&gt;&lt;br&gt;&lt;br&gt;
+\frac{(k + 1)(k + 2)(k + 3)(k + 4)} {4}</math><br><br>
 Since both the basis and the inductive step have been proved, it has now been proved by mathematical induction that S_n holds for all natural n
 [[introduction-TADM2E|Back to ''Introduction ...'']]