## anonymous one year ago A statement Sn about the positive integers is given. Write statements S1, S2, and S3, and show that each of these statements is true. Show your work. Sn: 12 + 42 + 72 + . . . + (3n - 2)2 = 𝑛(6𝑛2−3𝑛−1)2

1. anonymous

n(6n^2-3n-1)/2

2. anonymous

anybody got a clue

3. anonymous

@anonymous_user

4. phi

can you make this clearer (3n - 2)2 ?

5. anonymous

n(6n^2-3n-1)/2

6. phi

No, I mean the last term in the sum 12 + 42 + 72 + . . . + (3n - 2)2

7. anonymous

its 1^2+4^2+7^2

8. anonymous

im so lost

9. phi

oh. it looks like twelve + forty-two+... and I was not making any progress (who would have guessed)

10. anonymous

lol its my fualt i should have double checked thst sooo

11. phi

Do you have to prove the result or derive it? proving it should be easier

12. anonymous

no i just have to show thast the stamnets are true and show all my work proving them

13. phi

are you studying proof by induction?

14. anonymous

i think so

15. anonymous

yes iam

16. phi

the first step is show the identity is true for n=1 can you do that ?

17. anonymous

no im jsut now learning this stug=ff lol sorry im a newbie

18. phi

when n=1, what is "series" ?

19. phi

what is the last term of the summation when n=1 ?

20. anonymous

1

21. phi

and the starting term is 1 in other words the entire summation is 1^2 (just 1 term) on the right-hand side we have the formula n(6n^2-3n-1)/2 what is that when n=1 ?

22. anonymous

ummmmm

23. phi

you sub in n=1 into the formula n(6n^2-3n-1)/2 what do you get ?

24. anonymous

scratches head um n+1(6n^2-3n-1)/2

25. phi

the idea is everywhere you see n in the formula, erase it, and write in 1 in its place then simplify (now that you have a number instead of a letter, you can do that) what do you get ?

26. anonymous

1(61^2-31-1)/2

27. phi

ok. and remember when you have for example 3n that means 3*n ( multiply signs are usually left out) so the formula is 1*(6*1^2 - 3*1 -1)/2 now simplify that

28. anonymous

hold on im doing it now

29. anonymous

the answer is 1 right

30. phi

yes, it simplifies to (6-4)/2 = 2/2 = 1 that shows the formula works for n=1 because 1^2 = 1 (the answer we get from the formula)

31. phi

now the hard part. we assume the formula works for numbers up to n (whatever n happens to be) and show it also works for n+1 This part is difficult because it looks like we need some messy algebra.

32. anonymous

this is what i didi 1^2 evaluates to 1 Multiply 1 and 6 1 6*1^2 evaluates to 6 Multiply 1 and 3 1 3*1 evaluates to 3 6*1^2-3*1 evaluates to 3 6*1^2-3*1-1 evaluates to 2 Multiply 1 and 2 1 1*(6*1^2-3*1-1) evaluates to 2 1*(6*1^2-3*1-1)/2 evaluates to 1

33. anonymous

uhhh math

34. anonymous

so how dowe start this

35. phi

yes, that is how you evaluate it.

36. anonymous

sweet

37. phi

all of that proves that the formula works for n=1 now assume the formula works for numbers up to n show it works for n+1 Induction Step: Assume the identity is true for n: $1^2 + 4^2 + ... +(3n-1)^2 = \frac{n(6n^2-3n-1)}{2}$

38. phi

now add the next term to the summation that is add $$\left(3(n+1)-1 \right)^2$$ to both sides $1^2 + 4^2 + ... +(3n-1)^2+\left(3(n+1)-1 \right)^2 = \frac{n(6n^2-3n-1)}{2} +\left(3(n+1)-1 \right)^2$

39. anonymous

holy moly tat is hard

40. phi

yes, it is very difficult.

41. anonymous

so is there more too it plz say no lol

42. phi

the formula for n+1 becomes $\frac{(n+1)}{2} \left(6(n+1)^2 -3(n+1) -1\right)$ we have to show that $\frac{n(6n^2-3n-1)}{2} +\left(3(n+1)-1 \right)^2$ can be written as $\frac{(n+1)}{2} \left(6(n+1)^2 -3(n+1) -1\right)$

43. anonymous

thnx