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

Mathematics
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The formula for the nth term of an arithmetic sequence is a[n] = a[1]+(n-1)d. You know that d= a[k+1]-a[k] = 7, and a[1]=8, so the nth term is a[n] = 8 + 7(n-1)
ok 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...
a. a[n]=15+7n b. a[n]=1+7n c. a[n]=8+7n d. a[n]=1+7(n-1) e. a[n]= -1+8n

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If you expand the solution i sent, you'll find a[n] = 8 +7n - 7 = 1 + 7n...answer b.
oh oh ok sorry about that after I posted it I figured that out thank you and I am your fan now:)
No worries.
do you understand how to use the sigma notation?
Yes.
Use sigma notation to write the sum. (1/3.2)+(1/4.3)+...+(1/7.6)
a. \[\sum_{n=1}^{5}\] (1/(n+1)(n+2))
It will be sum from n=3 to 7 of (10/11n-1).
b. \[\sum_{n=1}^{5}(1/(n(n+1)))\]
ok thank you!
Do you need an explanation?
Yes that would be helpful... I mean I kinda get it but it wouldn't hurt anything
The denominator is in the form of some number n + (n-1)/10 (e.g. 3.2 = 3 + (3-1)/10). You then just take the reciprocal of n+(n-1)/10 and use sigma notation for n = 3 to 7.
I just cleaned up the fraction before using sigma.
oh ok makes it sound alot easier thanks!
\[\sum_{n=3}^{7}1/[n+(n-1)/10]\]
Find 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
I'm pretty sure the answer's -4/5, but I want to check something.
okay...
Yes...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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