Verifying the Equality for n = 2: Is the Equation Correct?

[tmt1]

I have this equality

$$\frac{2^n}{n!} + 1 = \frac{{2}^{n + 1}}{n + 1}$$

and I'm not sure if it is correct or how to verify if it is correct. I've tried a few different approaches that have led to dead ends.

 

[topsquark]

I have this equality

$$\frac{2^n}{n!} + 1 = \frac{{2}^{n + 1}}{n + 1}$$

and I'm not sure if it is correct or how to verify if it is correct. I've tried a few different approaches that have led to dead ends.

n = 1
\(\displaystyle \frac{2^1}{1!} + 1 = \frac{2}{1} + 1 = 2 + 1 = 3\)

\(\displaystyle \frac{2^{1 + 1}}{1 + 1} = \frac{2^2}{2} = \frac{4}{2} = 2\)

So the equation can only have solutions for certain n, not all.

Graphing it we can see that the equation has only two solutions close to, but not equal to 1. See below.

Graph

-Dan

 

[tmt1]

I have this equality

$$\frac{2^n}{n!} + 1 = \frac{{2}^{n + 1}}{n + 1}$$

and I'm not sure if it is correct or how to verify if it is correct. I've tried a few different approaches that have led to dead ends.

Sorry, I meant to write

$$\frac{2^n}{n!} + 1 = \frac{{2}^{n + 1}}{(n + 1)!}$$

Does this make more sense? I'm trying a bunch of ways to verify it, but none of it works.

 

[topsquark]

Sorry, I meant to write

$$\frac{2^n}{n!} + 1 = \frac{{2}^{n + 1}}{(n + 1)!}$$

Does this make more sense? I'm trying a bunch of ways to verify it, but none of it works.

n = 2

\(\displaystyle \frac{2^2}{2!} + 1 = \frac{4}{2} + 1 = 2 + 1 = 3\)

\(\displaystyle \frac{2^{2 + 1}}{(2 + 1)!} = \frac{2^3}{3!} = \frac{8}{6} = \frac{4}{3}\)

You really need to check these before you post them. (n = 1 doesn't work either. I chose n = 2 for variety.)

-Dan