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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.

Mathematics
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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
Im following

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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)
okay
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
this is the principle of mathematical induction. I didnt actually do your example yet, this just preliminary
I know, thank you, good notes
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
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?
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 ,
so we can rewrite the statement P(n)
ok im starting to understand
|dw:1366641535256:dw|
so we have here a left side and a right side of the statement we are trying to prove
P(1) : |dw:1366641602052:dw|
now the left side is just another way of saying, plug in 1 into the formula 2k+1 , no big deal.
I thought p(1) was 1(1+2)
we want to prove [2(1) +1] + [2(2) + 1] + [2(3)+1] + ... [2(n)+1] = n(n+1)
oh
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?
yes
because 3 = 2(1) + 1 5 = 2 (2) + 1 7 = 2(3) + 1
|dw:1366641946782:dw|
let's look at the case n=2 |dw:1366642018761:dw|
|dw:1366642104889:dw|
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?
yes, but which number is k because in my problem k isnt named
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)
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
oh dear i didnt copy correctly
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)
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.
3=3
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
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]
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
with me so far?
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]
yeah
using only algebra, basically
so how about we look at the conclusion for a minute
ok, let me do it another way, we can go back to this way later, it might be a little advanced
ok still with me?
yea Im following
we assume that [2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k)+ 1 ] = k ( k + 2) is true
from this hypothesis we want to deduce the conclusion [2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k+1)+ 1 ] = (k+1) [ (k+1) + 2]
proof: given [2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k)+ 1 ] = k ( k + 2) add [2(k+1) + 1 ] to both sides of equation
|dw:1366643507200:dw|
so we are just using algebra here
ok
you can add the same number to both sides of an equation, thats a legal algebraic move dfsd
ok im back , :)
ok
so we added [2(k+1) + 1 ] to both sides of the hypothesis , which was given
hello?
yes did you add it?
|dw:1366644222445:dw|
now the left side is what we want in our conclusion, and we can simplify the right side
okay
Im here but for some reason I keep losing connection
me too
i was lagged
me too
ok did you add both sides
I tried but my teacher said we already solved for what was needed.
[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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