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aussy123
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.
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
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)
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
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)
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?
because 3 = 2(1) + 1 5 = 2 (2) + 1 7 = 2(3) + 1
let's look at the case n=2 |dw:1366642018761: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.
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
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]
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
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
so we are just using algebra here
you can add the same number to both sides of an equation, thats a legal algebraic move dfsd
so we added [2(k+1) + 1 ] to both sides of the hypothesis , which was given
now the left side is what we want in our conclusion, and we can simplify the right side
Im here but for some reason I keep losing connection
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