A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
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
 3 years ago
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.

This Question is Closed

perl
 3 years ago
Best ResponseYou've already chosen the best response.1we 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

perl
 3 years ago
Best ResponseYou've already chosen the best response.1I am going to rewrite this as a proposition P(n) Let P(n) be the statement 3 + 5 + 7 + ... + 2n+1 = n(n+1)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I know, thank you, good notes

perl
 3 years ago
Best ResponseYou've already chosen the best response.1so 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
 3 years ago
Best ResponseYou've already chosen the best response.1ok 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
 3 years ago
Best ResponseYou've already chosen the best response.1notice 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
 3 years ago
Best ResponseYou've already chosen the best response.1so we can rewrite the statement P(n)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok im starting to understand

perl
 3 years ago
Best ResponseYou've already chosen the best response.1so we have here a left side and a right side of the statement we are trying to prove

perl
 3 years ago
Best ResponseYou've already chosen the best response.1P(1) : dw:1366641602052:dw

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I thought p(1) was 1(1+2)

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

perl
 3 years ago
Best ResponseYou've already chosen the best response.1ok 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?

perl
 3 years ago
Best ResponseYou've already chosen the best response.1because 3 = 2(1) + 1 5 = 2 (2) + 1 7 = 2(3) + 1

perl
 3 years ago
Best ResponseYou've already chosen the best response.1let's look at the case n=2 dw:1366642018761:dw

perl
 3 years ago
Best ResponseYou've already chosen the best response.1case 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
 3 years ago
Best ResponseYou've already chosen the best response.0yes, but which number is k because in my problem k isnt named

perl
 3 years ago
Best ResponseYou've already chosen the best response.1we 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
 3 years ago
Best ResponseYou've already chosen the best response.1the 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
 3 years ago
Best ResponseYou've already chosen the best response.1oh dear i didnt copy correctly

perl
 3 years ago
Best ResponseYou've already chosen the best response.1Correction 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
 3 years ago
Best ResponseYou've already chosen the best response.1Proof: 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.

perl
 3 years ago
Best ResponseYou've already chosen the best response.1next 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
 3 years ago
Best ResponseYou've already chosen the best response.1so 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
 3 years ago
Best ResponseYou've already chosen the best response.1i 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

perl
 3 years ago
Best ResponseYou've already chosen the best response.1so 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]

perl
 3 years ago
Best ResponseYou've already chosen the best response.1using only algebra, basically

perl
 3 years ago
Best ResponseYou've already chosen the best response.1so how about we look at the conclusion for a minute

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

perl
 3 years ago
Best ResponseYou've already chosen the best response.1we assume that [2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k)+ 1 ] = k ( k + 2) is true

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

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

perl
 3 years ago
Best ResponseYou've already chosen the best response.1so we are just using algebra here

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

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

perl
 3 years ago
Best ResponseYou've already chosen the best response.1now the left side is what we want in our conclusion, and we can simplify the right side

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Im here but for some reason I keep losing connection

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I tried but my teacher said we already solved for what was needed.

perl
 3 years ago
Best 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
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.