anonymous
  • anonymous
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^2-3n-1))/2
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
anonymous
  • anonymous
phi
  • phi
Is this two different questions?

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anonymous
  • anonymous
yes
phi
  • phi
The 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 ?
anonymous
  • anonymous
i got 40
phi
  • phi
I 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
  • anonymous
oh i thought you were talking about the first question (4(4n+1)(8n+7))/6 that
phi
  • phi
perhaps you should put in ^ to show exponents? \[ 1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = \frac{n(6n^2-3n-1)}{2} \]
phi
  • phi
that has a chance of being true
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
yea i got 1
phi
  • phi
if you don't use the equation editor, write it as 1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = (n(6n^2-3n-1))/2
anonymous
  • anonymous
ok so the second one is true, what about the first one, do i just show the work that its false
anonymous
  • anonymous
oh ok sorry im new to openstudy
phi
  • phi
for 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(6-3-1))/2 = 2/2 = 1 is also 1 so it works for n=1 does it work for n=2 ?
phi
  • phi
for n=2, on the left side we get 1^2 + 4^2 = 1+16= 17
anonymous
  • anonymous
its supposed to be 1^2 and 4^2 and 7^2 i just realised that sorry
anonymous
  • anonymous
wouldnt the left side include 7^2?
phi
  • phi
the 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, (3n-2)^2 is 1 and the left side is just 1 = right side stuff for n=2, we add 1 + (3*2-2)^2 or 1+ (6-2)^2 or 1 + 4^2 notice they showed the first 3 terms and then the formula (3n-2)^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
  • phi
or say it a different way 1^2 + 4^2 = (n(6n^2-3n-1))/2 where n=2
phi
  • phi
and it is worth checking that it is true before trying to prove it is true for any positive n
anonymous
  • anonymous
i got 17 for n2
anonymous
  • anonymous
ohhhh so if i were to try n3 then it would have to be equal to 1^2 + 4^2 + 7^2?
phi
  • phi
yes. 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
  • anonymous
yea it gave 66
anonymous
  • anonymous
the first one is false tho i just tested it
phi
  • phi
Induction proof begins by proving the formula works for the "first case" i.e. for n=1 in other words, show 1^2 = (n(6n^2-3n-1))/2 when n=1 we already did that, but make that the first part of the proof: Prove: \[ 1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = \frac{n(6n^2-3n-1)}{2} \] Proof by Induction. Base case: show the formula is true for n=1 \[ 1^2 = \frac{n(6n^2-3n-1)}{2} \text{with n=1} \\ 1= \frac{1 \cdot (6-3-1)}{2} \\ 1=1\] this proves the formula true for n=1
anonymous
  • anonymous
ok do i have to prove n2 also or is that good
phi
  • phi
Induction case: *assume* the formula is true for any positive n. Prove that it must also be true for n+1
anonymous
  • anonymous
i just got lost haha what did u do there?
phi
  • phi
We 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. \[ 1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 + (3(n+1)-2)^2 = \\ \frac{n(6n^2-3n-1)}{2} +(3(n+1)-2)^2 \]
phi
  • phi
now 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= k-1) and see if we can make the right side look like \[ \frac{k(6k^2-3k-1)}{2} \]
anonymous
  • anonymous
but why did u add \[(3(n+1)-2)^2\]
phi
  • phi
the 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
  • anonymous
ok so that is just to prove it works for the rest of the terms?
phi
  • phi
yes. 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
phi
  • phi
I assume you are lost?
anonymous
  • anonymous
a little bit but not that much
anonymous
  • anonymous
so if i wanted to prove it was true i just plug it in the original but just placing (n-1) for n
phi
  • phi
(n+1)
anonymous
  • anonymous
so what would be the final equation im pluging it into
phi
  • phi
The 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
  • phi
to make it as easy as possible, define k= n+1 (which makes n= k-1) replace n with k-1 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.
anonymous
  • anonymous
ok let me try
phi
  • phi
in other words, \[ \frac{n(6n^2-3n-1)}{2} +(3(n+1)-2)^2 \\ \text{replace n= k-1} \\ \frac{(k-1)(6(k-1)^2-3(k-1)-1)}{2} +(3k-2)^2 \] now simplify
phi
  • phi
It 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
  • anonymous
yea 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
  • anonymous
for both questions
phi
  • phi
yes.
anonymous
  • anonymous
ok and if n1 doesnt work, do i still have to replace it?
anonymous
  • anonymous
like do i still have to prove that it works for n+1
anonymous
  • anonymous
ok nevermind i got it, but could you help me with one more please
phi
  • phi
Here is an example of how you would write it up
anonymous
  • anonymous
for the given statement\[Pn\], write the statements \[P1, Pk,\] and \[Pk+1\] 2+4+6+.....+2n=n(n+1)
anonymous
  • anonymous
for the p1, pk and pk+1, the 1, k and k+1 are under the p
phi
  • phi
I 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
  • phi
for \(P_k\) just write down \(P_n\) but replace n with k for \(P_{k+1} \) put in k+1 for n
anonymous
  • anonymous
so for Pk, do i just say k(k+1)
anonymous
  • anonymous
or do i have to write down all the terms
phi
  • phi
just 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
  • anonymous
ok thank you! after i finish it can i tag you in another post and u can check if its correct
phi
  • phi
ok. I should be around, but it may take a few minutes.
anonymous
  • anonymous
ok
phi
  • phi
just to be clear **so for Pk, do i just say k(k+1)*** you of course write 2+4+...+2k = k(k+1)
anonymous
  • anonymous
ok

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