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Lukecrayonz
 4 years ago
Use sigma notation to write the sum:
[1(1/6)^2]+[1(2/6)^2]+...1(6/6)^2]
Lukecrayonz
 4 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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Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0Im not 100% sure, but i'm thinking its this?

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1330750567546:dw

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.0looks correct to me

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.0with the squared, since im really good at not noticing that

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.0\[ \sum_{n=1}^{6} (1(\frac{n}{6})^{2}) = (1(\frac{1}{6})^{2}) + (1(\frac{2}{6})^{2}) + ... + (1(\frac{6}{6})^2) \] :D

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0I'm trying to do a whole chapter in one night, so let's pull an all nighter?!:D

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0Write 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.

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.0i came online to find method of approximating square roots, snce i didnt want to pull out calculator. lol

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0a[1]=0.375, a[k+1]=a[k]+0.25

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.0seems 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

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0Here's an example answer, if this helps. http://screensnapr.com/v/2PVSpz.png

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.0a_(k+1) = a_1 + (k+1 1)d = a_1 + kd a_k = a_1 + (k1)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

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0I THINK it's a[2]=0.375+0.25=.625 a[3]=.625+0.25=0.875

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.0oh, i added wrong, 0.875 not using calculator

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0a[4]0.875+.25=1.125 a[5]1.125+.25=1.375

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0When you do a_k, do you mean sub k? Because when I put sub I put [k]

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.0yeah, when you use equation thing, subk is a_k, so that's why I use the _'s

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0So when I'm doing the nth term of the sequence and writing the function for it, what would that be?

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.0A(n) = a_1 + (n1)d; d = common difference, a_1 = first term would be the general form I'd use

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0Thank you! I think mine would work but yours seems more sophisticated.

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0Oh hey, your formula is the exact formula I'm using for my next questions.

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.0well, if you test n=1, .375 + .25 = .625, which was the second term

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.0yep, 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.

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0hmm, another question..

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0Find 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

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0I think the formula S[n]=n/2(a[1]+a[n]) would be used?

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0Just kidding far from that.

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0Don't even have a a[n]! Haha, what was I thinking :P

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.0my source with the formulas for this says the (a[1] + a[n] is in numerator

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0Here's an example answer, maybe this could clear up what the formula is..? http://screensnapr.com/v/zVwwkZ.png

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.0yeah, 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

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0http://screensnapr.com/v/x2890d.png But in that, where the hell did 9 come from?

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.0n1, n = 10 => 101 = 9

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0So i got a[10]=1.5+19(.5)=11

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.0yep, that's what I got as well

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0Then i got s[20]=10(1.5+11)

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.0Yep, got 125 as well. :)

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0BOO! REAL WORLD APPLICATION PROBLEMS!

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.015*16*17*18*19*20*21? Haha

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0Adding not multiplying, that would be an incredible amount of logs.

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0Uhh, a[1]=15, a[6]=21 and s[6]=21/2(15+21)=126

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.0i 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

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.0but it does appear that 126 is correct.
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