Algebraic manipulation with factorials

[NP04]

Homework Statement

If 𝑎𝑛 =𝑥𝑛2𝑛𝑛! , find 𝑎𝑛+1𝑎𝑛 . See picture.
Edit by mentor:
If $a_n = \frac{x^n}{2^n n!}$, find $\frac {a_{n+1}}{a_n}$

Relevant Equations

Substitution

I substituted and got ((x_n/2ⁿn!) + 1)/(x_n/2ⁿn!). I then multiplied by 2ⁿn! to each side and got (x_n + 2ⁿn!)/(x_n). Now I am confused as to what my next step should be.

 

[Mark44]

Problem Statement: If 𝑎𝑛 =𝑥𝑛2𝑛𝑛! , find 𝑎𝑛+1𝑎𝑛 . See picture.
Relevant Equations: Substitution

View attachment 247799

I substituted and got ((x_n/2ⁿn!) + 1)/(x_n/2ⁿn!). I then multiplied by 2ⁿn! to each side and got (x_n + 2ⁿn!)/(x_n). Now I am confused as to what my next step should be.

You can simplify $\frac{x^n}{2^n}$ to $\left(\frac x 2\right)^n$. When you are calculating $\frac{a_{n+1}}{a_n}$, you'll have a fraction divided by a fraction. Keep in mind that $\frac{a/b}{c/d}$ is the same as $\frac a b\cdot \frac d c$

 

[RPinPA]

I substituted and got $((x_n/2^n n!) + 1)$

Wait, you're saying that the $(n+1)$-th term is found by adding 1 to the $n$-th term? Are you sure about that?

You might want to check a few terms. Evaluate $a_1$, $a_2$ and $a_3$ and see if you think that's still true.

 

[archaic]

What you found is wrong, here's what you did :
$$\frac{\frac{x^n}{2^nn!}+1}{\frac{x^n}{2^nn!}}=\frac{(a_n)+1}{a_n}$$
To put it simply, you interpreted $a_{n+1}$ as $(a_n)+1$. The way you should look at $a_n$ is $a=f(n)$.
Let's take, for example, $b_n=2n$, this means that $b$ is a function of $n$, or $b=f(n)=2n$. Should we try to find $b_{n+1}$, what we'll do is plug $n+1$ in $f$, namely we'll compute $f(n+1)$ which is $f(n+1)=2(n+1)=2n+2$.