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aussy123
Group Title
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.
 one year ago
 one year ago
aussy123 Group Title
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.
 one year ago
 one year ago

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perl Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

aussy123 Group TitleBest ResponseYou've already chosen the best response.0
Im following
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
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)
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.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
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
this is the principle of mathematical induction. I didnt actually do your example yet, this just preliminary
 one year ago

aussy123 Group TitleBest ResponseYou've already chosen the best response.0
I know, thank you, good notes
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
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 ,
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
so we can rewrite the statement P(n)
 one year ago

aussy123 Group TitleBest ResponseYou've already chosen the best response.0
ok im starting to understand
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
dw:1366641535256:dw
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
so we have here a left side and a right side of the statement we are trying to prove
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
P(1) : dw:1366641602052:dw
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
now the left side is just another way of saying, plug in 1 into the formula 2k+1 , no big deal.
 one year ago

aussy123 Group TitleBest ResponseYou've already chosen the best response.0
I thought p(1) was 1(1+2)
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
we want to prove [2(1) +1] + [2(2) + 1] + [2(3)+1] + ... [2(n)+1] = n(n+1)
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.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?
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
because 3 = 2(1) + 1 5 = 2 (2) + 1 7 = 2(3) + 1
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
dw:1366641946782:dw
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
let's look at the case n=2 dw:1366642018761:dw
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
dw:1366642104889:dw
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

aussy123 Group TitleBest ResponseYou've already chosen the best response.0
yes, but which number is k because in my problem k isnt named
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
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)
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.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
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
oh dear i didnt copy correctly
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
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)
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.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]
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
i didnt do the proof yet , just showing you what we want to prove. we are proving what is called a 'conditional' statement. an 'ifthen' statement
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
with me so far?
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
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]
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
using only algebra, basically
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
so how about we look at the conclusion for a minute
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
ok, let me do it another way, we can go back to this way later, it might be a little advanced
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
ok still with me?
 one year ago

aussy123 Group TitleBest ResponseYou've already chosen the best response.0
yea Im following
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
we assume that [2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k)+ 1 ] = k ( k + 2) is true
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
from this hypothesis we want to deduce the conclusion [2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k+1)+ 1 ] = (k+1) [ (k+1) + 2]
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
proof: given [2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k)+ 1 ] = k ( k + 2) add [2(k+1) + 1 ] to both sides of equation
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
dw:1366643507200:dw
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
so we are just using algebra here
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
you can add the same number to both sides of an equation, thats a legal algebraic move dfsd
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
ok im back , :)
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
so we added [2(k+1) + 1 ] to both sides of the hypothesis , which was given
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
yes did you add it?
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
dw:1366644222445:dw
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
now the left side is what we want in our conclusion, and we can simplify the right side
 one year ago

aussy123 Group TitleBest ResponseYou've already chosen the best response.0
Im here but for some reason I keep losing connection
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
ok did you add both sides
 one year ago

aussy123 Group TitleBest ResponseYou've already chosen the best response.0
I tried but my teacher said we already solved for what was needed.
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.1
[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
 one year ago
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