Find the formula for the sequence $\frac{5}{1},\frac{8}{2},\frac{11}{6},\frac{14}{24},\frac{17}{120}$

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I'm supposed to find the formula for the sequence:

$$\frac{5}{1},\frac{8}{2},\frac{11}{6},\frac{14}{24},\frac{17}{120}$$

I started by writing this:

$$(\frac{1}{n!})$$

But I don't know how to get the numerator. I know there's a difference of 3 between the numbers, but I don't know how to approach creating a formula for it.

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2 Answers

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So you see it starts with 5 for $n=1$ and roses by 3 every time. So $5+3(n-1)$ will be the numerator. Simplify everything and put it together with the denominator. $(2+3n)/n!$

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Notice the numerator increases by $3$ each time and so the common difference is $d=3$. Now use the formula for the arithmetic sequence $a_n=a+(n-1)d$, where $a$ is the first term and $d$ the common difference. Thus the numerator has the form $5+(n-1)3=2+3n$. And the denominator is as you said $n!$. Thus the sequence is $\{a_n\}_{n=1}^{\infty}$ where$$a_n=\frac{2+3n}{n!}$$

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