Given $e^{-5}\frac{5^n}{n!}=e^{-5}\frac{5^{n+1}}{(n+1)!},$ find the value of $n$

Question

I do not know how to find the value of $n$ this is a past paper question I am trying to solve, any help is much appreciated. Thank you.

Answer 1

Multiply both sides by $\frac{e^5(n+1)!}{5^n}$ simplify and get

$$5=n+1\rightarrow n=4$$

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how do you know you must multiply both sides by...

I do not know. It is just a conclusion. To realize that you can rewrite you equation in the following way:

$$\frac{e^{-5}\cdot 5^n}{n!}=\frac{e^{-5}\cdot 5^n\cdot 5}{(n+1)\cdot n!}$$

As you can see, both sides contain

$$\frac{e^{-5}\cdot5^n}{n!}$$

thus this eliminates and it remains only

$$1=\frac{5}{n+1}$$

thus evidently, $n=4$

Answer 2

A variant: simplify the common factors: $$\mathrm e^{-5}\frac{5^n}{n!}=\mathrm e^{-5}\frac{5^n\cdot 5}{n!\,(n+1)}\iff 1=\frac 5{n+1}\iff n+1=5.$$