Ask your own question, for FREE!
Mathematics 19 Online
OpenStudy (anonymous):

series question

OpenStudy (anonymous):

http://gyazo.com/11d360e31e398a1174844702893f4562

OpenStudy (anonymous):

how do i start?

OpenStudy (solomonzelman):

Plug in 1 for n, for next term plug in 2 for n And so forth. Good?

OpenStudy (anonymous):

\[a_n = (3-1(2^{-1}) + (3-2(2^{-2}) +...+ (3-n(2^{-n})\] thats all?

OpenStudy (solomonzelman):

Nothing more, and then find the sum of the series.

OpenStudy (solomonzelman):

I mean you put in pluses already, but there is an easier way to find the sum of the series. Write out the series and I'll show, OK?

OpenStudy (anonymous):

is it a geometric series?

OpenStudy (solomonzelman):

Write it out, and will see.

OpenStudy (anonymous):

(2.5 +2.5 +2.625 + ...)

OpenStudy (solomonzelman):

Can you please put it in fractions?

OpenStudy (anonymous):

( 5/2 + 5/2 + 21/8 +...)

OpenStudy (solomonzelman):

The series is infinite, so if the terms are increasing you can't find the sum

OpenStudy (solomonzelman):

Are they?

OpenStudy (anonymous):

i have no idea

OpenStudy (solomonzelman):

OK, looking at the series, your exponents are becoming negative and therefore the bigger the n the bigger the fraction is going to be formed, Now look at the problem closely keeping what I told you in mind. Would you agree that the bigger it is going to get the more it is going to approach 3? (they are not asking for exact sum, b/c it doesn't exist in infinite series.

OpenStudy (solomonzelman):

@mathsnerd101, do you see why is the series ('s sum) going to approach 3 the bigger the series gets?

OpenStudy (solomonzelman):

So the answer is 3.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!