TADM2E 1.11 - Revision history

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The basis case is when <math>n = 0</math> 
:<math>\sum_{i=1}^0 i^2 = 0^2 = 0 </math> 
and using
<math>n=0</math>
in the formula
<math>\frac {n(n + 1)(2 \cdot n + 1)} {6}</math>
you get: 
:<math>\frac {0(0 + 1)(2 \cdot 0 + 1)} {6} = \frac {0(1)(0 + 1)} {6} = \frac {0} {6} = 0</math> 
Since these are equal, the basis case is true. 

Now, I will show that on the assumption that the summation is true for ''n'', it follows that it is true for <math>n+1</math>

:<math>\sum_{i = 1}^{n + 1} i^2 =
(n + 1)^2 + \sum_{i = 1}^n i^2 =
(n^2 + 2n + 1) + \frac{n(n + 1)(2 \cdot n + 1)} {6} =
\frac{6n^2 + 12n + 6}{6}+ \frac{2n^3 + 3n^2 + n} {6} =
\frac {2n^3 + 9n^2 + 13n + 6} {6}</math> 

Which should be equal to the formula
<math>\frac {n(n + 1)(2 \cdot n + 1)} {6}</math>
when <math>n = n + 1</math>:
<math>\frac {(n + 1)(n + 1 + 1)(2 \cdot (n + 1) + 1)} {6}=
\frac {(n + 1)(n + 2)(2n + 2 + 1)} {6}=
\frac {(n^2 + 3n + 2)(2n + 3)} {6}=
\frac {2n^3 + 9n^2 + 13n + 6} {6}</math>

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

← Older revision		Revision as of 09:42, 31 July 2020
Line 1:		Line 1:
−	~~The basis case is when <math>n = 0</math><br>~~
−	~~:<math>\sum_{i=1}^0 i^2 = 0^2 = 0 </math><br>~~
−	~~and using~~
−	~~<math>n=0</math>~~
−	~~in the formula~~
−	~~<math>\frac {n(n + 1)(2 \cdot n + 1)} {6}</math>~~
−	~~you get:<br>~~
−	~~:<math>\frac {0(0 + 1)(2 \cdot 0 + 1)} {6} = \frac {0(1)(0 + 1)} {6} = \frac {0} {6} = 0</math><br>~~
−	~~Since these are equal, the basis case is true.<br>~~

−
−	~~Now, I will show that on the assumption that the summation is true for ''n'', it follows that it is true for <math>n+1</math>~~
−
−	~~:<math>\sum_{i = 1}^{n + 1} i^2 =~~
−	~~(n + 1)^2 + \sum_{i = 1}^n i^2 =~~
−	~~(n^2 + 2n + 1) + \frac{n(n + 1)(2 \cdot n + 1)} {6} =~~
−	~~\frac{6n^2 + 12n + 6}{6}+ \frac{2n^3 + 3n^2 + n} {6} =~~
−	~~\frac {2n^3 + 9n^2 + 13n + 6} {6}</math><br>~~
−
−	~~Which should be equal to the formula~~
−	~~<math>\frac {n(n + 1)(2 \cdot n + 1)} {6}</math>~~
−	~~when <math>n = n + 1</math>:~~
−	~~<math>\frac {(n + 1)(n + 1 + 1)(2 \cdot (n + 1) + 1)} {6}=~~
−	~~\frac {(n + 1)(n + 2)(2n + 2 + 1)} {6}=~~
−	~~\frac {(n^2 + 3n + 2)(2n + 3)} {6}=~~
−	~~\frac {2n^3 + 9n^2 + 13n + 6} {6}</math>~~
−
−
−	~~[[introduction-TADM2E\|Back to ''Introduction ...'']]~~

@@ Line 1: / Line 1: @@
-The basis case is when &lt;math&gt;n = 0&lt;/math&gt;&lt;br&gt;
+The basis case is when <math>n = 0</math><br>
-:&lt;math&gt;\sum_{i=1}^0 i^2 = 0^2 = 0 &lt;/math&gt;&lt;br&gt;
+:<math>\sum_{i=1}^0 i^2 = 0^2 = 0 </math><br>
 and using
-&lt;math&gt;n=0&lt;/math&gt;
+<math>n=0</math>
 in the formula
-&lt;math&gt;\frac {n(n + 1)(2 \cdot n + 1)} {6}&lt;/math&gt;
+<math>\frac {n(n + 1)(2 \cdot n + 1)} {6}</math>
-you get:&lt;br&gt;
+you get:<br>
-:&lt;math&gt;\frac {0(0 + 1)(2 \cdot 0 + 1)} {6} = \frac {0(1)(0 + 1)} {6} = \frac {0} {6} = 0&lt;/math&gt;&lt;br&gt;
+:<math>\frac {0(0 + 1)(2 \cdot 0 + 1)} {6} = \frac {0(1)(0 + 1)} {6} = \frac {0} {6} = 0</math><br>
-Since these are equal, the basis case is true.&lt;br&gt;
+Since these are equal, the basis case is true.<br>
-Now, I will show that on the assumption that the summation is true for ''n'', it follows that it is true for &lt;math&gt;n+1&lt;/math&gt;
+Now, I will show that on the assumption that the summation is true for ''n'', it follows that it is true for <math>n+1</math>
-:&lt;math&gt;\sum_{i = 1}^{n + 1} i^2 =
+:<math>\sum_{i = 1}^{n + 1} i^2 =
 (n + 1)^2 + \sum_{i = 1}^n i^2 =
 (n^2 + 2n + 1) + \frac{n(n + 1)(2 \cdot n + 1)} {6} =
 \frac{6n^2 + 12n + 6}{6}+ \frac{2n^3 + 3n^2 + n} {6} =
-\frac {2n^3 + 9n^2 + 13n + 6} {6}&lt;/math&gt;&lt;br&gt;
+\frac {2n^3 + 9n^2 + 13n + 6} {6}</math><br>
 Which should be equal to the formula
-&lt;math&gt;\frac {n(n + 1)(2 \cdot n + 1)} {6}&lt;/math&gt;
+<math>\frac {n(n + 1)(2 \cdot n + 1)} {6}</math>
-when &lt;math&gt;n = n + 1&lt;/math&gt;:
+when <math>n = n + 1</math>:
-&lt;math&gt;\frac {(n + 1)(n + 1 + 1)(2 \cdot (n + 1) + 1)} {6}=
+<math>\frac {(n + 1)(n + 1 + 1)(2 \cdot (n + 1) + 1)} {6}=
 \frac {(n + 1)(n + 2)(2n + 2 + 1)} {6}=
 \frac {(n^2 + 3n + 2)(2n + 3)} {6}=
-\frac {2n^3 + 9n^2 + 13n + 6} {6}&lt;/math&gt;
+\frac {2n^3 + 9n^2 + 13n + 6} {6}</math>
 [[introduction-TADM2E|Back to ''Introduction ...'']]

← Older revision	Revision as of 00:55, 1 August 2020
Line 1:	Line 1:
	+	The basis case is when <math>n = 0</math><br>
	+	:<math>\sum_{i=1}^0 i^2 = 0^2 = 0 </math><br>
	+	and using
	+	<math>n=0</math>
	+	in the formula
	+	<math>\frac {n(n + 1)(2 \cdot n + 1)} {6}</math>
	+	you get:<br>
	+	:<math>\frac {0(0 + 1)(2 \cdot 0 + 1)} {6} = \frac {0(1)(0 + 1)} {6} = \frac {0} {6} = 0</math><br>
	+	Since these are equal, the basis case is true.<br>

	+
	+	Now, I will show that on the assumption that the summation is true for ''n'', it follows that it is true for <math>n+1</math>
	+
	+	:<math>\sum_{i = 1}^{n + 1} i^2 =
	+	(n + 1)^2 + \sum_{i = 1}^n i^2 =
	+	(n^2 + 2n + 1) + \frac{n(n + 1)(2 \cdot n + 1)} {6} =
	+	\frac{6n^2 + 12n + 6}{6}+ \frac{2n^3 + 3n^2 + n} {6} =
	+	\frac {2n^3 + 9n^2 + 13n + 6} {6}</math><br>
	+
	+	Which should be equal to the formula
	+	<math>\frac {n(n + 1)(2 \cdot n + 1)} {6}</math>
	+	when <math>n = n + 1</math>:
	+	<math>\frac {(n + 1)(n + 1 + 1)(2 \cdot (n + 1) + 1)} {6}=
	+	\frac {(n + 1)(n + 2)(2n + 2 + 1)} {6}=
	+	\frac {(n^2 + 3n + 2)(2n + 3)} {6}=
	+	\frac {2n^3 + 9n^2 + 13n + 6} {6}</math>
	+
	+
	+	[[introduction-TADM2E\|Back to ''Introduction ...'']]