- 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

- chestercat

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- anonymous

- anonymous

- phi

Is this two different questions?

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## More answers

- anonymous

yes

- 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

i got 40

- 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

oh i thought you were talking about the first question (4(4n+1)(8n+7))/6 that

- 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

that has a chance of being true

- anonymous

oh ok

- anonymous

yea i got 1

- 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

ok so the second one is true, what about the first one, do i just show the work that its false

- anonymous

oh ok sorry im new to openstudy

- 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

for n=2, on the left side we get
1^2 + 4^2 = 1+16= 17

- anonymous

its supposed to be 1^2 and 4^2 and 7^2 i just realised that sorry

- anonymous

wouldnt the left side include 7^2?

- 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

or say it a different way
1^2 + 4^2 = (n(6n^2-3n-1))/2
where n=2

- phi

and it is worth checking that it is true before trying to prove it is true for any positive n

- anonymous

i got 17 for n2

- anonymous

ohhhh so if i were to try n3 then it would have to be equal to 1^2 + 4^2 + 7^2?

- 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

yea it gave 66

- anonymous

the first one is false tho i just tested it

- 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

ok do i have to prove n2 also or is that good

- phi

Induction case: *assume* the formula is true for any positive n. Prove that it must also be true for n+1

- anonymous

i just got lost haha what did u do there?

- 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

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

but why did u add \[(3(n+1)-2)^2\]

- 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

ok so that is just to prove it works for the rest of the terms?

- 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

I assume you are lost?

- anonymous

a little bit but not that much

- 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

(n+1)

- anonymous

so what would be the final equation im pluging it into

- 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

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

ok let me try

- 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

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

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

for both questions

- phi

yes.

- anonymous

ok and if n1 doesnt work, do i still have to replace it?

- anonymous

like do i still have to prove that it works for n+1

- anonymous

ok nevermind i got it, but could you help me with one more please

- phi

Here is an example of how you would write it up

##### 1 Attachment

- anonymous

for the given statement\[Pn\], write the statements \[P1, Pk,\] and \[Pk+1\]
2+4+6+.....+2n=n(n+1)

- anonymous

for the p1, pk and pk+1, the 1, k and k+1 are under the p

- 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

for \(P_k\) just write down \(P_n\) but replace n with k
for \(P_{k+1} \) put in k+1 for n

- anonymous

so for Pk, do i just say k(k+1)

- anonymous

or do i have to write down all the terms

- 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

ok thank you! after i finish it can i tag you in another post and u can check if its correct

- phi

ok. I should be around, but it may take a few minutes.

- anonymous

ok

- 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

ok

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