anonymous
  • anonymous
Find the sum of each finite geometric series. 1+2+4+...+2048
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
kropot72
  • kropot72
The common ratio r is 2 in this case. When the common ratio is positive and greater than 1 the formula for the sum of n terms is \[S _{n}=\frac{a(r ^{n}-1)}{r-1}\] For this question we need to find the value of n for the last term 2048. The nth term is found from the formula \[n ^{th} term=ar ^{n-1}\] Plugging in the known values of a and r gives \[2048=1\times2^{n-1}\ .........(1)\] 2 to the power of 11 = 2048. Therefore n = 11 - 1 = 10 Now you just need to plug the values of a, r and n into the formula for the sum of n terms.
anonymous
  • anonymous
the last part is the part I don't understand, can you explain it simply what I have to do "to plug in"
kropot72
  • kropot72
Sorry, my bad. The final part of my post should be: "2 to the power of 11 = 2048. Therefore n = 11 + 1 = 12. Now you just need to plug the values of a, r and n into the formula for the sum of n terms." The values are as follows: a = 1 r = 2 n = 12 The formula is \[S _{n}=\frac{a(r ^{n}-1)}{r-1}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I thought it was n-1 so wouldn't it be 11 - 1 instead of 11 + 1
kropot72
  • kropot72
\[2^{11}=2048\] \[2^{n-1}=2048\] Therefore n - 1 = 11 ........(2) Adding 1 to both sides of equation (2) gives n = 11 + 1 = 12
anonymous
  • anonymous
oh okay!
anonymous
  • anonymous
can you work me through the formula?
kropot72
  • kropot72
The values are as follows: a = 1 r = 2 n = 12 The formula is \[S _{n}=\frac{a(r ^{n}-1)}{r-1}\] So after plugging in the values we get \[S _{12}=\frac{1(2^{12}-1)}{2-1}=you\ can\ calculate\]
anonymous
  • anonymous
that's actually the part I don't get, don't understand how to calculate it :(
anonymous
  • anonymous
2^18 = 262144?
kropot72
  • kropot72
Where did you get '2 to the power of 18' from?
anonymous
  • anonymous
oh gosh, I misread it.
anonymous
  • anonymous
oh gosh, I misread it.
anonymous
  • anonymous
the 12 looked like an 18 to me, I'm sorry, haha
anonymous
  • anonymous
so 4096
kropot72
  • kropot72
When you simplify the fraction you get \[S _{12}=\frac{1(2^{12}-1)}{2-1}=2^{12}-1\]
anonymous
  • anonymous
4095 is it then?
kropot72
  • kropot72
4095 is correct!
anonymous
  • anonymous
thank you so much! I actually think I'm starting to get it, I just need nudges in the right direction! :)
kropot72
  • kropot72
You're welcome :) Glad to hear you are understanding.

Looking for something else?

Not the answer you are looking for? Search for more explanations.