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anonymous
 one year ago
Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.
(4 points each.)
4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) = (4(4n+1)(8n+7))/6
12 + 42 + 72 + ... + (3n  2)2 = (n(6n^23n1))/2
anonymous
 one year ago
Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false. (4 points each.) 4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) = (4(4n+1)(8n+7))/6 12 + 42 + 72 + ... + (3n  2)2 = (n(6n^23n1))/2

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phi
 one year ago
Best ResponseYou've already chosen the best response.1Is this two different questions?

phi
 one year ago
Best ResponseYou've already chosen the best response.1The second equation does not make sense. we must show true for all positive integers n, so n=1 should work put in n=1 into the last equation. what do you get ?

phi
 one year ago
Best ResponseYou've already chosen the best response.1I assume the last equation's last term is \[ (3n  2)^2 \] with n=1 , you get \[ (3  2)^2 = 1^2 =1 \] and that should be the first term in the series. but they show it starting with 12.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh i thought you were talking about the first question (4(4n+1)(8n+7))/6 that

phi
 one year ago
Best ResponseYou've already chosen the best response.1perhaps you should put in ^ to show exponents? \[ 1^2 + 4^2 + 7^2 + ... + (3n  2)^2 = \frac{n(6n^23n1)}{2} \]

phi
 one year ago
Best ResponseYou've already chosen the best response.1that has a chance of being true

phi
 one year ago
Best ResponseYou've already chosen the best response.1if you don't use the equation editor, write it as 1^2 + 4^2 + 7^2 + ... + (3n  2)^2 = (n(6n^23n1))/2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so the second one is true, what about the first one, do i just show the work that its false

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh ok sorry im new to openstudy

phi
 one year ago
Best ResponseYou've already chosen the best response.1for induction. show the base case is true (and it does not hurt to try the next few n, just in case it does not really work) for n=1 the left side is 1 and the right side (1(631))/2 = 2/2 = 1 is also 1 so it works for n=1 does it work for n=2 ?

phi
 one year ago
Best ResponseYou've already chosen the best response.1for n=2, on the left side we get 1^2 + 4^2 = 1+16= 17

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its supposed to be 1^2 and 4^2 and 7^2 i just realised that sorry

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wouldnt the left side include 7^2?

phi
 one year ago
Best ResponseYou've already chosen the best response.1the formula on the left shows a sum of terms each term is defined as (3n  2)^2 where n=1, then 2, then 3, etc. up to some final "N" for n=1, (3n2)^2 is 1 and the left side is just 1 = right side stuff for n=2, we add 1 + (3*22)^2 or 1+ (62)^2 or 1 + 4^2 notice they showed the first 3 terms and then the formula (3n2)^2 for the "nth term" but I want to just test the first 2 terms, so we go from n=1 up to n=2

phi
 one year ago
Best ResponseYou've already chosen the best response.1or say it a different way 1^2 + 4^2 = (n(6n^23n1))/2 where n=2

phi
 one year ago
Best ResponseYou've already chosen the best response.1and it is worth checking that it is true before trying to prove it is true for any positive n

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohhhh so if i were to try n3 then it would have to be equal to 1^2 + 4^2 + 7^2?

phi
 one year ago
Best ResponseYou've already chosen the best response.1yes. so it looks like the formula works for n=2 (17 on both sides) for n=3 the left side is 66, so the formula on the right (with n=3) should give 66 It probably does (but you can double check). time to prove it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the first one is false tho i just tested it

phi
 one year ago
Best ResponseYou've already chosen the best response.1Induction proof begins by proving the formula works for the "first case" i.e. for n=1 in other words, show 1^2 = (n(6n^23n1))/2 when n=1 we already did that, but make that the first part of the proof: Prove: \[ 1^2 + 4^2 + 7^2 + ... + (3n  2)^2 = \frac{n(6n^23n1)}{2} \] Proof by Induction. Base case: show the formula is true for n=1 \[ 1^2 = \frac{n(6n^23n1)}{2} \text{with n=1} \\ 1= \frac{1 \cdot (631)}{2} \\ 1=1\] this proves the formula true for n=1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok do i have to prove n2 also or is that good

phi
 one year ago
Best ResponseYou've already chosen the best response.1Induction case: *assume* the formula is true for any positive n. Prove that it must also be true for n+1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i just got lost haha what did u do there?

phi
 one year ago
Best ResponseYou've already chosen the best response.1We write down the series for all the terms up to n, and then add the next term that means the very next term we replace n with (n+1) . then we add that same messy term to the right side. \[ 1^2 + 4^2 + 7^2 + ... + (3n  2)^2 + (3(n+1)2)^2 = \\ \frac{n(6n^23n1)}{2} +(3(n+1)2)^2 \]

phi
 one year ago
Best ResponseYou've already chosen the best response.1now the messy part. we want to show the formula works for (n+1) to make the algebra easier(??), I would let k= n+1 (so n= k1) and see if we can make the right side look like \[ \frac{k(6k^23k1)}{2} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but why did u add \[(3(n+1)2)^2\]

phi
 one year ago
Best ResponseYou've already chosen the best response.1the idea is we assume the formula is true for all terms up to n now add the next term (which is (n+1) put into the formula (3n  2)^2 i.e. (3(n+1)2)^2 if we add that term to the left side, we also must add it to the right side. if we label k=n+1, then the formula on the right side should simplify to the same formula we started with but with k rather than n

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so that is just to prove it works for the rest of the terms?

phi
 one year ago
Best ResponseYou've already chosen the best response.1yes. induction relies on 1) proving it works for n=1 2) proving if it works for terms up to n , then it works for n+1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a little bit but not that much

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so if i wanted to prove it was true i just plug it in the original but just placing (n1) for n

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so what would be the final equation im pluging it into

phi
 one year ago
Best ResponseYou've already chosen the best response.1The logic goes like this: assume the formula works up to "n" we write down the series on the left = the formula on the right for n terms then we add the (n+1) term to both sides. By assumption this must still be true. now "simplify" the right side as much as possible. If we get the original formula (but with (n+1) in place of everywhere we originally had n, this shows the formula works.

phi
 one year ago
Best ResponseYou've already chosen the best response.1to make it as easy as possible, define k= n+1 (which makes n= k1) replace n with k1 in the right hand side simplify if the formula simplifies to the original formula with k instead of n, then we have shown the formula works for the n+1 term.

phi
 one year ago
Best ResponseYou've already chosen the best response.1in other words, \[ \frac{n(6n^23n1)}{2} +(3(n+1)2)^2 \\ \text{replace n= k1} \\ \frac{(k1)(6(k1)^23(k1)1)}{2} +(3k2)^2 \] now simplify

phi
 one year ago
Best ResponseYou've already chosen the best response.1It has taken me about 3 tries to get it to simplify down to k (k^2 3k 1)/2 (which proves the formula works for the n+1 term) It is very messy algebra!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea i can tell, so all i do to prove is first prove n1 then prove that it workes when u substitute with n+1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok and if n1 doesnt work, do i still have to replace it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0like do i still have to prove that it works for n+1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok nevermind i got it, but could you help me with one more please

phi
 one year ago
Best ResponseYou've already chosen the best response.1Here is an example of how you would write it up

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for the given statement\[Pn\], write the statements \[P1, Pk,\] and \[Pk+1\] 2+4+6+.....+2n=n(n+1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for the p1, pk and pk+1, the 1, k and k+1 are under the p

phi
 one year ago
Best ResponseYou've already chosen the best response.1I assume it looks something like this: \[ P_n : 2+4+6+.....+2n=n(n+1) \] and you want \(P_1\). Replace n with 1 and you get 1*(1+1)= 2 on the right side on the left side, the last term is 2n= 2*1= 2. Because the series starts at 2, and the lasat term is 2, we have just 1 term thus \[ P_1 : 2 = 2 \]

phi
 one year ago
Best ResponseYou've already chosen the best response.1for \(P_k\) just write down \(P_n\) but replace n with k for \(P_{k+1} \) put in k+1 for n

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so for Pk, do i just say k(k+1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or do i have to write down all the terms

phi
 one year ago
Best ResponseYou've already chosen the best response.1just k(k+1) unless you have a specific number, you can only write the "idea" (and even if you know k (say k=100) you would not write all the terms. Just the general formula)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok thank you! after i finish it can i tag you in another post and u can check if its correct

phi
 one year ago
Best ResponseYou've already chosen the best response.1ok. I should be around, but it may take a few minutes.

phi
 one year ago
Best ResponseYou've already chosen the best response.1just to be clear **so for Pk, do i just say k(k+1)*** you of course write 2+4+...+2k = k(k+1)
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