Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

wcaprarBest ResponseYou've already chosen the best response.0
Determine the smallest number of terms required to approximate the sum of the series with an error less than 0.0001. \[\sum_{n}^{\infty}(7^k)/(6^kk!)\]
 2 years ago

wcaprarBest ResponseYou've already chosen the best response.0
I understand that you would set the sum equal to .0001 and solve for k. Right?
 2 years ago

wcaprarBest ResponseYou've already chosen the best response.0
I can simplify the equation by saying (1/k!)x(7/6)^k right?
 2 years ago

bmpBest ResponseYou've already chosen the best response.1
If I understood the problem correctly, there are a couple of ways to do it. Remember that the remainder estimate for alternating series is given by:\[R_n \le b_{n+1}\]So yeah, you can simplify it and put in the form:\[0,0001 \le \frac{7^{k+1}}{6^{k+1}(k+1)!} = \frac{1}{(k+1)!}*(\frac{7}{6})^{k+1}\]Or if you have a calculator, you can start expanding some terms until one of them is < 0.0001
 2 years ago

wcaprarBest ResponseYou've already chosen the best response.0
I did that and I got to the seventh term, it was the right answer! Thanks! But on the test I won't be able to use a calculator, can this problem be solved some way w/o one?
 2 years ago

bmpBest ResponseYou've already chosen the best response.1
If they ask for the smallest number, I guess you have to 1) manually compute the values or 2) solve an equation of the form above. If they do not, just pick a reasonable number and compute it, like n = 10. Very likely the error will be really small and it would suffice the boundary for Rn
 2 years ago

bmpBest ResponseYou've already chosen the best response.1
And by the way, doing a series test without a calculator sucks a lot, huh
 2 years ago

eliassaabBest ResponseYou've already chosen the best response.0
I agree with @bmp and add that you must fire your teacher.
 2 years ago

wcaprarBest ResponseYou've already chosen the best response.0
Im not sure if we will be given this same type of problem but we did go over a problem like this in class that was WAY easier that we solved with very little effort
 2 years ago

eliassaabBest ResponseYou've already chosen the best response.0
Why should one compute 7^k for k=1 to 7 and similar problems by hand. Those antiquated teachers should be trained to undersatand that the world has changed and they should follow the flow.
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.