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## Lukecrayonz 3 years ago Use sigma notation to write the sum: [1-(1/6)^2]+[1-(2/6)^2]+...1-(6/6)^2]

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1. Lukecrayonz

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

2. Lukecrayonz

|dw:1330750567546:dw|

3. AccessDenied

looks correct to me

4. Hunus

That's correct.

5. mathg8

good work

6. AccessDenied

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

7. AccessDenied

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

8. Lukecrayonz

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

9. Lukecrayonz

Uhh, im stuck D:

10. Lukecrayonz

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

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

12. Lukecrayonz

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

13. AccessDenied

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

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

15. AccessDenied

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

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

17. AccessDenied

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

18. Lukecrayonz

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

19. Lukecrayonz

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

20. AccessDenied

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

21. Lukecrayonz

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

22. Lukecrayonz

a[n].375n+.25?

23. AccessDenied

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

24. Lukecrayonz

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

25. Lukecrayonz

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

26. Lukecrayonz

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

27. AccessDenied

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

28. Lukecrayonz

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

29. AccessDenied

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

hmm, another question..

31. Lukecrayonz

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

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

33. Lukecrayonz

Just kidding far from that.

34. Lukecrayonz

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

35. AccessDenied

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

36. Lukecrayonz

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

37. AccessDenied

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

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

39. Lukecrayonz

n-1?

40. AccessDenied

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

41. Lukecrayonz

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

42. Lukecrayonz

a[20] sorry

43. AccessDenied

yep, that's what I got as well

44. Lukecrayonz

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

45. Lukecrayonz

which equals 125

46. AccessDenied

Yep, got 125 as well. :)

47. Lukecrayonz

BOO! REAL WORLD APPLICATION PROBLEMS!

48. Lukecrayonz
49. Lukecrayonz

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

50. Lukecrayonz

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

51. Lukecrayonz

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

52. AccessDenied

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

but it does appear that 126 is correct.

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