## 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^2-3n-1))/2

1. anonymous

@misssunshinexxoxo

2. anonymous

@phi

3. phi

Is this two different questions?

4. anonymous

yes

5. 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 ?

6. anonymous

i got 40

7. 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.

8. anonymous

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

9. phi

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

10. phi

that has a chance of being true

11. anonymous

oh ok

12. anonymous

yea i got 1

13. 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

14. anonymous

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

15. anonymous

oh ok sorry im new to openstudy

16. 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 ?

17. phi

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

18. anonymous

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

19. anonymous

wouldnt the left side include 7^2?

20. 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

21. phi

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

22. phi

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

23. anonymous

i got 17 for n2

24. anonymous

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

25. 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.

26. anonymous

yea it gave 66

27. anonymous

the first one is false tho i just tested it

28. 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

29. anonymous

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

30. phi

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

31. anonymous

i just got lost haha what did u do there?

32. 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. $1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 + (3(n+1)-2)^2 = \\ \frac{n(6n^2-3n-1)}{2} +(3(n+1)-2)^2$

33. 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 $\frac{k(6k^2-3k-1)}{2}$

34. anonymous

but why did u add $(3(n+1)-2)^2$

35. 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

36. anonymous

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

37. 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

38. phi

I assume you are lost?

39. anonymous

a little bit but not that much

40. anonymous

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

41. phi

(n+1)

42. anonymous

so what would be the final equation im pluging it into

43. 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.

44. 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.

45. anonymous

ok let me try

46. 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

47. 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!

48. 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?

49. anonymous

for both questions

50. phi

yes.

51. anonymous

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

52. anonymous

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

53. anonymous

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

54. phi

Here is an example of how you would write it up

55. anonymous

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

56. anonymous

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

57. 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 $P_1 : 2 = 2$

58. phi

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

59. anonymous

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

60. anonymous

or do i have to write down all the terms

61. 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)

62. anonymous

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

63. phi

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

64. anonymous

ok

65. 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)

66. anonymous

ok