Question

can anyone simplify this?
n(2x+3)^n+1/(n+2)*(n+1)/(n-1)(2x+3)^n

Answer 1

\[\left| \frac{ n(2x+3)^{n+1}(n+1) }{ (n+2)(n-1)(2x+3)^{n} } \right|\] I'm guessing you're looking for some sort of radius of convergence or something and this was a ratio test. Well, if so, then we can simplify this only a little bit, but we don't need to simplify it much anyway.
\[\left| \frac{ n(2x+3)^{n+1}(n+1) }{ (n+2)(n-1)(2x+3)^{n} } \right| = \left| \frac{ n(n+1)(2x+3) }{ (n+2)(n-1) } \right| \]
\[= \left| 2x+3 \right|\left| \frac{ n(n+1) }{ (n+2)(n-1) } \right|\]
Now if we want this to converge, we want to take lim as n-> infinity and find x such that |2x+3| < 1. So we have
\[\left| 2x+3 \right|*\lim_{n \rightarrow \infty}\left| \frac{ n(n+1) }{ (n+2)(n-1) } \right| < 1\]
\[|2x+3|(1) < 1\] I figure you could do the rest from here, but as long as you can dumb it down just enough to isolate some reasonable expression of x, the rest isnt difficult.

Answer 2

Hey, thank you! I eventually figured it out on khan academy, but they have very limited examples and often don't show how they obtain answers in the "hints". So it was tedious. I really appreciate the help as there isn't many people on here who answer calculus questions like this.