Question

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

Answer 1

@misssunshinexxoxo

Answer 2

@phi

Answer 3

Is this two different questions?

Answer 4

yes

Answer 5

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 ?

Answer 6

i got 40

Answer 7

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.

Answer 8

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

Answer 9

perhaps you should put in ^ to show exponents?
\[ 1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = \frac{n(6n^2-3n-1)}{2} \]

Answer 10

that has a chance of being true

Answer 11

oh ok

Answer 12

yea i got 1

Answer 13

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

Answer 14

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

Answer 15

oh ok sorry im new to openstudy

Answer 16

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 ?

Answer 17

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

Answer 18

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

Answer 19

wouldnt the left side include 7^2?

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

i got 17 for n2

Answer 24

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

Answer 25

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.

Answer 26

yea it gave 66

Answer 27

the first one is false tho i just tested it

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

i just got lost haha what did u do there?

Answer 32

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.
\[ 1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 + (3(n+1)-2)^2 = \\
\frac{n(6n^2-3n-1)}{2} +(3(n+1)-2)^2 \]

Answer 33

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
\[ \frac{k(6k^2-3k-1)}{2} \]

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

I assume you are lost?

Answer 39

a little bit but not that much

Answer 40

so if i wanted to prove it was true i just plug it in the original but just placing (n-1) for n

Answer 41

(n+1)

Answer 42

so what would be the final equation im pluging it into

Answer 43

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.

Answer 44

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.

Answer 45

ok let me try

Answer 46

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

Answer 47

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!

Answer 48

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?

Answer 49

for both questions

Answer 50

yes.

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

Here is an example of how you would write it up

Answer 55

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

Answer 56

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

Answer 57

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
\[ P_1 : 2 = 2 \]

Answer 58

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

Answer 59

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

Answer 60

or do i have to write down all the terms

Answer 61

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)

Answer 62

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

Answer 63

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

Answer 64

ok

Answer 65

just to be clear
**so for Pk, do i just say k(k+1)***

you of course write 2+4+...+2k = k(k+1)

Answer 66

ok