Approximate the sum of the series correct to four decimal places.
sum n=1,infinity ((-1)^(n-1) n^2)/(10^n)
I know how to finish the equation once I have the number I need to go up to, but I don't know how to find that number, can anyone help?
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
You can do the same thing with this one, just solve another equality without the (-1)^(n-1) n^2/(10^n)<0.0001
This should give you the number of terms
Not the answer you are looking for? Search for more explanations.
It wants me to find the sum.
I think the easiest way to do this one is write out the terms, maybe the first 3 or 4. and then find the first term that is less than 0.0001.
Add all of the terms before this term and you get the approximate sum correct to 0.0001
(cause i just realized in the equality was like 10^n which you can't really solve for easily )