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anonymous
 5 years ago
Write the nth term of the arithmetic sequence as a function of n.
a[1]=8, a[k+1]=a[k]+7
anonymous
 5 years ago
Write the nth term of the arithmetic sequence as a function of n. a[1]=8, a[k+1]=a[k]+7

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The formula for the nth term of an arithmetic sequence is a[n] = a[1]+(n1)d. You know that d= a[k+1]a[k] = 7, and a[1]=8, so the nth term is a[n] = 8 + 7(n1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so I was close but the problem is that none of the answers I can choose from match with that...let me show you my choices...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a. a[n]=15+7n b. a[n]=1+7n c. a[n]=8+7n d. a[n]=1+7(n1) e. a[n]= 1+8n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you expand the solution i sent, you'll find a[n] = 8 +7n  7 = 1 + 7n...answer b.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh oh ok sorry about that after I posted it I figured that out thank you and I am your fan now:)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you understand how to use the sigma notation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Use sigma notation to write the sum. (1/3.2)+(1/4.3)+...+(1/7.6)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a. \[\sum_{n=1}^{5}\] (1/(n+1)(n+2))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It will be sum from n=3 to 7 of (10/11n1).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0b. \[\sum_{n=1}^{5}(1/(n(n+1)))\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you need an explanation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes that would be helpful... I mean I kinda get it but it wouldn't hurt anything

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The denominator is in the form of some number n + (n1)/10 (e.g. 3.2 = 3 + (31)/10). You then just take the reciprocal of n+(n1)/10 and use sigma notation for n = 3 to 7.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I just cleaned up the fraction before using sigma.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok makes it sound alot easier thanks!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=3}^{7}1/[n+(n1)/10]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Find the sum of the infinite series. \[\sum_{i=1}^{\infty}4(1/4)^i\] a.undefined b.4/3 c.4 d.8/5 e.4/5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm pretty sure the answer's 4/5, but I want to check something.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes...I don't know what level of maths you're doing, but first you need to ensure that the series is convergent. Since it's absolutely convergent, it will be convergent. It also means you can split the sum up into the positive and negative components and use the formula for the limiting value in a geometric series to determine each of the sums. You then end up with 4/5. Do you know how to do this?
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