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Lukecrayonz
Use sigma notation to write the sum: [1-(1/6)^2]+[1-(2/6)^2]+...1-(6/6)^2]
Im not 100% sure, but i'm thinking its this?
|dw:1330750567546:dw|
looks correct to me
with the squared, since im really good at not noticing that
\[ \sum_{n=1}^{6} (1-(\frac{n}{6})^{2}) = (1-(\frac{1}{6})^{2}) + (1-(\frac{2}{6})^{2}) + ... + (1-(\frac{6}{6})^2) \] :D
I'm trying to do a whole chapter in one night, so let's pull an all nighter?!:D
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.
i came online to find method of approximating square roots, snce i didnt want to pull out calculator. lol
a[1]=0.375, a[k+1]=a[k]+0.25
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
Here's an example answer, if this helps. http://screensnapr.com/v/2PVSpz.png
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
I THINK it's a[2]=0.375+0.25=.625 a[3]=.625+0.25=0.875
oh, i added wrong, 0.875 -not using calculator-
a[4]0.875+.25=1.125 a[5]1.125+.25=1.375
When you do a_k, do you mean sub k? Because when I put sub I put [k]
yeah, when you use equation thing, sub-k is a_k, so that's why I use the _'s
So when I'm doing the nth term of the sequence and writing the function for it, what would that be?
A(n) = a_1 + (n-1)d; d = common difference, a_1 = first term would be the general form I'd use
Thank you! I think mine would work but yours seems more sophisticated.
Oh hey, your formula is the exact formula I'm using for my next questions.
well, if you test n=1, .375 + .25 = .625, which was the second term
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.
hmm, another question..
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
I think the formula S[n]=n/2(a[1]+a[n]) would be used?
Just kidding far from that.
Don't even have a a[n]! Haha, what was I thinking :P
my source with the formulas for this says the (a[1] + a[n] is in numerator
Here's an example answer, maybe this could clear up what the formula is..? http://screensnapr.com/v/zVwwkZ.png
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
http://screensnapr.com/v/x2890d.png But in that, where the hell did 9 come from?
n-1, n = 10 => 10-1 = 9
So i got a[10]=1.5+19(.5)=11
yep, that's what I got as well
Then i got s[20]=10(1.5+11)
Yep, got 125 as well. :)
BOO! REAL WORLD APPLICATION PROBLEMS!
15*16*17*18*19*20*21? Haha
Adding not multiplying, that would be an incredible amount of logs.
Uhh, a[1]=15, a[6]=21 and s[6]=21/2(15+21)=126
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
but it does appear that 126 is correct.