Recent content by jaxx

oh it's just subtraction, I see. Thanks very much for all your help then, I really appreciate it.

so for 99(-2)^n - 96 (-2)^n, since 3 factors into both 99 and 96, you can simply make it 3(-2n)?

Alright thanks, my only question now is what it means to "combine like terms", and how you get 3 from 2^n(99 - 96)

just out of curiosity, was I right when I got to this? -96 = - 99 / (n - 1) x [-2^(n-1)] if so, am I right by multiplying by -2 to get rid of the (^n - 1), and ending up with -192 = (99 / n - 1) (-2n), dividing by -2, getting 96 = (99 / n - 1) x n? I don't know what to do from there but was I...

Hence: t1[1−(-2)n]=99We have: t1−t1(-2)n=99why did it become t1 - t1? where did that come from?also, how did you just get rid of "n" at the end? there was never a determined = t1 equation, and when you substituted [1] that had n in it as well, so how did it disappear?

So why did you multiply by -1 = 1/-1? Can you not just multiply by 1/-1? Not sure I understand that bit, and multiplying by 2 / -2 gets rid of the -1 in (n - 1)? Also don't you have to multiply both sides of an equation by the same thing? Even though it is essentially the same but the negatives...

I guess I just don't understand at all why you would multiply both sides of t1(−2)n−1=48 by -2 / t1, and much less how to actually work through it from there. I get why your way is easier, and it finds t1 instead of taking the extra step to find n and then produce t1, but if possible, would you...

Right, I get the first part now. I get to t1 = -99 / -2^n - 1, and then substitute it into the tn formula like you did, which makes 48 = (-99 / -2^n - 1) [-2^(n-1)], and then multiply by -2 because it looks like that's what you did to get rid of it in the denominator. So I get -96 = - 99 / (n...

sorry I'm not sure how you got to t1 = 99 / 1 - (-2)^n. I get to 33 = t1[1 - (-2)^n] / 3, then wouldn't you multiply by 3 to get rid of the denominator, making it 99 = t1[1 - (-2)^n]? either way I see you did multiply by 3, but I'm not sure what you did to flip the equation to get 1 - (-2)^n...

I need to find the value of the first term for this geometric series. Sn = 33 tn = 48 r = -2 I know that I have to take the formulas tn = t1 x r^(n-1), and Sn = [t1 x (r^n) - 1] / (r - 1), and isolate t1 for the first formula and then input that into the second, but I don't know the actual...