PLEASE! IM BEGGING!! HELP ME!! PLEASE!! *TEARS*
Can someone please help me with this? I post it earlier but the answer wasnt what my teacher wanted. Can someone help me I will post in attachment.

- anonymous

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- anonymous

##### 1 Attachment

- perl

we want to prove that the formula
3 + 5 + 7 + ... + 2n+1 = n(n+1) is true for positive integers n=1,2,3,...
the problem is, there are an infinite number of statements to check, and we will never get done, one for each number n.
so we use mathematical induction

- anonymous

Im following

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## More answers

- perl

I am going to rewrite this as a proposition P(n)
Let P(n) be the statement 3 + 5 + 7 + ... + 2n+1 = n(n+1)

- anonymous

okay

- perl

mathematical induction says :
If you have a statement that is true for some number, for example n=1
(this is called the basis case)
and furthermore that it can be proven if P(k) is true, then P(k+1) is true.
then it must be the case that P(n) is true for all n greater than your basis

- perl

this is the principle of mathematical induction.
I didnt actually do your example yet, this just preliminary

- anonymous

I know, thank you, good notes

- perl

so for instance if your basis case P(1), and assuming you showed P(k)->P(k+1)
then you have P(2) because P(1) is true and P(1) ->P(2) is true
then it must be true P(2) is true , etc

- perl

ok thats just the logic behind it.
so in our example P(n) : 3 + 5 + 7 + ... + 2n+1 = n(n+1)
is P(1) true?

- perl

notice on the left side, you are summing a sequence of odd numbers 3,5,7,...The formula for the sequence of odd numbers is 2n+1 ,

- perl

so we can rewrite the statement P(n)

- anonymous

ok im starting to understand

- perl

|dw:1366641535256:dw|

- perl

so we have here a left side and a right side of the statement we are trying to prove

- perl

P(1) : |dw:1366641602052:dw|

- perl

now the left side is just another way of saying, plug in 1 into the formula 2k+1 , no big deal.

- anonymous

I thought p(1) was 1(1+2)

- perl

we want to prove
[2(1) +1] + [2(2) + 1] + [2(3)+1] + ... [2(n)+1] = n(n+1)

- anonymous

oh

- perl

ok let me start over, we want to prove this statement is true for all n.
[2(1) +1] + [2(2) + 1] + [2(3)+1] + ... [2(n)+1] = n(n+1)
do you agree?

- anonymous

yes

- perl

because 3 = 2(1) + 1
5 = 2 (2) + 1
7 = 2(3) + 1

- perl

|dw:1366641946782:dw|

- perl

let's look at the case n=2 |dw:1366642018761:dw|

- perl

|dw:1366642104889:dw|

- perl

case n=1
2(1) + 1 = 1 (1+1)
case n=2
[ 2(1) + 1 ] + [2 (2) + 1 ] = 2 (2+1)
case n=3
[ 2(1) + 1 ] + [2 (2) + 1 ] + [ 2(3) + 1] = 3 (3+1)
do you start to see a pattern?

- anonymous

yes, but which number is k because in my problem k isnt named

- perl

we want to prove all these statements are true without actually have to check each one by hand or calculator
[2(1) + 1] = 1 (1+1)
[ 2(1) + 1 ] + [2 (2) + 1 ] = 2 (2+1)
[ 2(1) + 1 ] + [2 (2) + 1 ] + [ 2(3) + 1] = 3 (3+1)
[ 2(1) + 1 ] + [2 (2) + 1 ] + [ 2(3) + 1] +[2(4) + 1] = 4 (4+1)
.
.
.
[2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(n)+ 1 ] = n ( n + 1)

- perl

the way to prove all of the statements, first show it is true for the first one
[2(1) + 1] = 1 (1+1) , does the left side equal right side (yes it does).
thats the basis case

- perl

oh dear i didnt copy correctly

- perl

Correction
we want to prove all these statements are true for any number n without actually have to check each one by hand or calculator
[2(1) + 1] = 1 (1+2)
[ 2(1) + 1 ] + [2 (2) + 1 ] = 2 (2+2)
[ 2(1) + 1 ] + [2 (2) + 1 ] + [ 2(3) + 1] = 3 (3+2)
[ 2(1) + 1 ] + [2 (2) + 1 ] + [ 2(3) + 1] +[2(4) + 1] = 4 (4+2)
.
.
.
[2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(n)+ 1 ] = n ( n + 2)

- perl

Proof: first show it is true for the first case
[2(1) + 1] = 1 (1+2) ( Check to see if the left side equals right side (yes it does).
This is called the basis case.

- anonymous

3=3

- perl

next step, (called the inductive step)
show that if the statement is true for n=k, then it will be true for n=k+1

- perl

so we prove that if this is true :
[2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k)+ 1 ] = k ( k + 2)
Then this must be true
[2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k+1)+ 1 ] = (k+1) [ (k+1) + 2]

- perl

i didnt do the proof yet , just showing you what we want to prove.
we are proving what is called a 'conditional' statement.
an 'if-then' statement

- perl

with me so far?

- perl

so we are given the 'if' part of the condition, also called the inductive hypothesis.
so we assume that it is true that
[2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k)+ 1 ] = k ( k + 2)
now how can we go from this statement to
this statement
[2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k+1)+ 1 ] = (k+1) [ (k+1) + 2]

- anonymous

yeah

- perl

using only algebra, basically

- perl

so how about we look at the conclusion for a minute

- perl

ok, let me do it another way, we can go back to this way later, it might be a little advanced

- perl

ok still with me?

- anonymous

yea Im following

- perl

we assume that
[2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k)+ 1 ] = k ( k + 2) is true

- perl

from this hypothesis we want to deduce the conclusion
[2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k+1)+ 1 ] = (k+1) [ (k+1) + 2]

- perl

proof:
given
[2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k)+ 1 ] = k ( k + 2)
add [2(k+1) + 1 ] to both sides of equation

- perl

|dw:1366643507200:dw|

- perl

so we are just using algebra here

- anonymous

ok

- perl

you can add the same number to both sides of an equation, thats a legal algebraic move
dfsd

- perl

ok im back , :)

- anonymous

ok

- perl

so we added [2(k+1) + 1 ] to both sides of the hypothesis , which was given

- anonymous

hello?

- perl

yes did you add it?

- perl

|dw:1366644222445:dw|

- perl

now the left side is what we want in our conclusion, and we can simplify the right side

- anonymous

okay

- anonymous

Im here but for some reason I keep losing connection

- perl

me too

- perl

i was lagged

- anonymous

me too

- perl

ok did you add both sides

- anonymous

I tried but my teacher said we already solved for what was needed.

- perl

[2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k)+ 1 ] = k ( k + 2)
+[2(k+1) +1 ] + [2(k+1)+1]
---------------------------------------------------------
[2(1)+1] +[2(2)+1] +... +[2(k)+1] +[2(k+1)+1] = k(k+2) + [2(k+1)+1

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