Talk:Introduction-TADM2E - Revision history

Matt at 18:11, 3 May 2020

2020-05-03T18:11:12Z

Matt: /* Solution 1.8 Missing */

2020-02-17T16:47:52Z

‎Solution 1.8 Missing

Matt: /* Solution 1.8 Missing */ new section

2020-02-17T16:47:00Z

‎Solution 1.8 Missing: new section

Parth: back to main page link added

2015-10-09T09:07:40Z

back to main page link added

Parth: Solution for TADM2E 1.10 , 1.12 and 1.14

2015-10-09T08:58:07Z

Solution for TADM2E 1.10 , 1.12 and 1.14

Parth: Solution for TADM2E 1.10 and 1.12

2015-10-09T08:18:53Z

Solution for TADM2E 1.10 and 1.12

Parth: Solution for 1-10

2015-10-09T05:53:47Z

Solution for 1-10

New page

= TADM2E 1.10 =

First we'll verify the base case for <math>n = 1</math> :

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

Now, we'll assume given statement is true up to <math>n - 1</math>.

So,

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

To prove for the general case <math>n</math>,

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

← Older revision		Revision as of 18:11, 3 May 2020
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	--[[User:Parth\|Parth]] ([[User talk:Parth\|talk]]) 05:07, 9 October 2015 (EDT)		--[[User:Parth\|Parth]] ([[User talk:Parth\|talk]]) 05:07, 9 October 2015 (EDT)
−
−	~~== TADM2E 1.8 ==~~
−
−	It looks like the solution to problem 1-8 was never written. I even tried going to http://algorist.com/algowiki/index.php/TADM2E_1.8, which doesn't exist. I would add it but, of course, I'm looking at this wiki specifically because I have no idea what I'm doing.

@@ Line 81: / Line 81: @@
 --[[User:Parth|Parth]] ([[User talk:Parth|talk]]) 05:07, 9 October 2015 (EDT)
-== Solution 1.8 Missing ==
+== TADM2E 1.8 ==
 It looks like the solution to problem 1-8 was never written. I even tried going to http://algorist.com/algowiki/index.php/TADM2E_1.8, which doesn't exist. I would add it but, of course, I'm looking at this wiki specifically because I have no idea what I'm doing.

← Older revision		Revision as of 16:47, 17 February 2020
Line 80:		Line 80:

	--[[User:Parth\|Parth]] ([[User talk:Parth\|talk]]) 05:07, 9 October 2015 (EDT)		--[[User:Parth\|Parth]] ([[User talk:Parth\|talk]]) 05:07, 9 October 2015 (EDT)
		+
		+	== Solution 1.8 Missing ==
		+
		+	It looks like the solution to problem 1-8 was never written. I even tried going to http://algorist.com/algowiki/index.php/TADM2E_1.8, which doesn't exist. I would add it but, of course, I'm looking at this wiki specifically because I have no idea what I'm doing.

@@ Line 18: / Line 18: @@
 : <math>\sum_{i=1}^n i = \sum_{i=1}^{n-1} i + n = \frac{(n-1)n}{2} + n = \frac{n^2 - n + 2n}{2} = \frac{n^2 + n}{2} = \frac{n(n+1)}{2}</math>
 ----
@@ Line 44: / Line 47: @@
 </math>
 ----
@@ Line 70: / Line 76: @@
 Thus,
 :<math>\sum_{i=0}^n a^i = \frac{a^{n+1}-1}{a-1}</math>

← Older revision		Revision as of 08:58, 9 October 2015
Line 43:		Line 43:
	\end{align}		\end{align}
	</math>		</math>
		+
		+	----
		+
		+	== TADM2E 1.14 ==
		+
		+	First we'll verify the base case for <math> n = 1</math> and we also assume <math>a=2</math>
		+
		+	:<math>\sum_{i=0}^1 a^i = (2)^0 + (2)^1 = \frac{(2)^{1+1} - 1}{2 - 1} = 3</math>
		+
		+	Now, we'll assume given statement is true up to <math>n - 1</math>.<br>
		+	So, following statement is true.
		+
		+	:<math>\sum_{i=0}^{n-1} a^i = \frac{a^{n-1+1}-1}{a-1} = \frac{a^n-1}{a-1}</math>
		+
		+	To prove for the general case <math>n</math>,
		+
		+	:<math>
		+	\begin{align}
		+	\sum_{i=0}^n a^i & = \sum_{i=0}^{n-1} a^i + a^n \\
		+	& = \frac{a^n-1}{a-1} + a^n \\
		+	& = \frac{a^n-1+a^n(a-1)}{a-1} \\
		+	& = \frac{a^n-1+a^{n+1}-a^n}{a-1} \\
		+	\end{align}
		+	</math>
		+
		+	Thus,
		+	:<math>\sum_{i=0}^n a^i = \frac{a^{n+1}-1}{a-1}</math>

@@ Line 1: / Line 1: @@
-= TADM2E 1.10 =
+As some even number problem's solution has already been provided so please find solution for following
 First we'll verify the base case for <math>n = 1</math> :
@@ Line 14: / Line 17: @@
 : <math>\sum_{i=1}^n i = \sum_{i=1}^{n-1} i + n = \frac{(n-1)n}{2} + n = \frac{n^2 - n + 2n}{2} = \frac{n^2 + n}{2} = \frac{n(n+1)}{2}</math>