## Lukecrayonz Group Title Use sigma notation to write the sum: [1-(1/6)^2]+[1-(2/6)^2]+...1-(6/6)^2] 2 years ago 2 years ago

1. Lukecrayonz Group Title

Im not 100% sure, but i'm thinking its this?

2. Lukecrayonz Group Title

|dw:1330750567546:dw|

3. AccessDenied Group Title

looks correct to me

4. Hunus Group Title

That's correct.

5. mathg8 Group Title

good work

6. AccessDenied Group Title

with the squared, since im really good at not noticing that

7. AccessDenied Group Title

$\sum_{n=1}^{6} (1-(\frac{n}{6})^{2}) = (1-(\frac{1}{6})^{2}) + (1-(\frac{2}{6})^{2}) + ... + (1-(\frac{6}{6})^2)$ :D

8. Lukecrayonz Group Title

I'm trying to do a whole chapter in one night, so let's pull an all nighter?!:D

9. Lukecrayonz Group Title

Uhh, im stuck D:

10. Lukecrayonz Group Title

Write the first five terms of the arithmetic sequence. Find the common difference and write the nth term of the sequence as a function of n.

11. AccessDenied Group Title

i came online to find method of approximating square roots, snce i didnt want to pull out calculator. lol

12. Lukecrayonz Group Title

a[1]=0.375, a[k+1]=a[k]+0.25

13. AccessDenied Group Title

seems like the common difference would be 0.25 since k and k+1 are successive terms of the sequence and we add 0.25 to a(k) to get a(k+1), i gotta check that again to make sure -- it's been a while since i did this stuff. ;x

14. Lukecrayonz Group Title

Here's an example answer, if this helps. http://screensnapr.com/v/2PVSpz.png

15. AccessDenied Group Title

a_(k+1) = a_1 + (k+1 -1)d = a_1 + kd a_k = a_1 + (k-1)d = a_1 + kd - d = (a_(k+1)) - d --> a_(k+1) = a_1 + kd - d + d = (a_k) + d yeah, sounds about right 0.25 being the common difference, i'd just add 0.25 to the first term five times 0.375 + 0.25 = 0.625 0.625 + 0.25 = 0.845 etc

16. Lukecrayonz Group Title

I THINK it's a[2]=0.375+0.25=.625 a[3]=.625+0.25=0.875

17. AccessDenied Group Title

oh, i added wrong, 0.875 -not using calculator-

18. Lukecrayonz Group Title

a[4]0.875+.25=1.125 a[5]1.125+.25=1.375

19. Lukecrayonz Group Title

When you do a_k, do you mean sub k? Because when I put sub I put [k]

20. AccessDenied Group Title

yeah, when you use equation thing, sub-k is a_k, so that's why I use the _'s

21. Lukecrayonz Group Title

So when I'm doing the nth term of the sequence and writing the function for it, what would that be?

22. Lukecrayonz Group Title

a[n].375n+.25?

23. AccessDenied Group Title

A(n) = a_1 + (n-1)d; d = common difference, a_1 = first term would be the general form I'd use

24. Lukecrayonz Group Title

Thank you! I think mine would work but yours seems more sophisticated.

25. Lukecrayonz Group Title

Oh hey, your formula is the exact formula I'm using for my next questions.

26. Lukecrayonz Group Title

a[1]=0, d=-2/3

27. AccessDenied Group Title

well, if you test n=1, .375 + .25 = .625, which was the second term

28. Lukecrayonz Group Title

0+(n-1)(-2/3)

29. AccessDenied Group Title

yep, looks correct there. :D I don't know of a more intuitive way to remember the formula except just using it often, since I actually forgot it since going over it in Algebra, though.

30. Lukecrayonz Group Title

hmm, another question..

31. Lukecrayonz Group Title

Find the nth partial sum of the arithmetic series, a[1]=1.5 d=.5 n=20 If it matters, the series is 1.5, 1.45, 1.40, 1.35

32. Lukecrayonz Group Title

I think the formula S[n]=n/2(a[1]+a[n]) would be used?

33. Lukecrayonz Group Title

Just kidding far from that.

34. Lukecrayonz Group Title

Don't even have a a[n]! Haha, what was I thinking :P

35. AccessDenied Group Title

my source with the formulas for this says the (a[1] + a[n] is in numerator

36. Lukecrayonz Group Title

Here's an example answer, maybe this could clear up what the formula is..? http://screensnapr.com/v/zVwwkZ.png

37. AccessDenied Group Title

yeah, that looks accurate to the formula I found $S_n = \frac{n(a_1 + a_n)}{2}$ I guess it wants you to find the arithmetic series equation, substitute in n=20 to it, and then use the formula for the nth partial sum

38. Lukecrayonz Group Title

http://screensnapr.com/v/x2890d.png But in that, where the hell did 9 come from?

39. Lukecrayonz Group Title

n-1?

40. AccessDenied Group Title

n-1, n = 10 => 10-1 = 9

41. Lukecrayonz Group Title

So i got a[10]=1.5+19(.5)=11

42. Lukecrayonz Group Title

a[20] sorry

43. AccessDenied Group Title

yep, that's what I got as well

44. Lukecrayonz Group Title

Then i got s[20]=10(1.5+11)

45. Lukecrayonz Group Title

which equals 125

46. AccessDenied Group Title

Yep, got 125 as well. :)

47. Lukecrayonz Group Title

BOO! REAL WORLD APPLICATION PROBLEMS!

48. Lukecrayonz Group Title
49. Lukecrayonz Group Title

15*16*17*18*19*20*21? Haha

50. Lukecrayonz Group Title

Adding not multiplying, that would be an incredible amount of logs.

51. Lukecrayonz Group Title

Uhh, a[1]=15, a[6]=21 and s[6]=21/2(15+21)=126

52. AccessDenied Group Title

i think the seventh term should be 21 since there are seven rows, with the first row being 15, sevond is 16, third is 17, fourth = 18, 5th = 19, 6th = 20, and 7th = 21

53. AccessDenied Group Title

but it does appear that 126 is correct.