I used to (and still do in a lot of ways) have problems with series. I find them terribly abstract, but I did finally "see" it, so keep at it and you will get there!(2.) Find the sum of the following series -
-3/(2^2) + 6/(2^5) - 9/(2^8).....
I know about the geometric addition but I honestly don't know how to solve this one...
Sometimes it helps to go ahead and multiply out your exponents to see what kind of mulitples you are working with:
First, since you have alternating signs, we know that this will be an alternating series, in otherwords, some form of will be part of your series. Next, notice that the denominators are multiplied by 8 to get the next denominator in the series, and that the exponents in the denominators are one number less than the numbers in their numerators.
So, usually it is best to start at n=0 to see if that will work with your series:
In your numerator you have multiples of 3. 3 x 1, 3 x 2, 3 x 2,... If you start at n=0, you can easily see that your numerator of your series will be 3n. Now in the denominator, it might help you to write out the 2's in long form: i.e. (2 x 2), (2 x 2 x 2 x 2 x 2), (2 x 2 x 2 x 2 x 2 x 2 x 2 x 2), ....
So we have , , , ... The exponents are one less than the numbers in the numerators, right? So, we have:
I'm not sure what your question is here. This looks like a logistic equation differential equation.(3.) My last question is regarding logistic differential equations. If we want to find a shortcut of the numerical value of a logistic equation as it goes to infinity, we take shortcuts by spotting the value of n in differential equations like dp/dt = kp(1-p/n) and dp/dt = (k/n)p(n-p). Here's my question, what would lim f'(t) (t to infinity) be? Is it the same as n or what?