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

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The equation under discussion, $$\frac{2^n}{n!} + 1 = \frac{{2}^{n + 1}}{(n + 1)!}$$, is being verified for various values of n. For n = 1, the left side equals 3 while the right side equals 2, indicating the equation does not hold. Similarly, for n = 2, the left side results in 3, while the right side equals 4/3, further confirming the equation's inconsistency. Graphical analysis suggests that the equation has limited solutions, specifically near but not equal to 1. The discussion emphasizes the need for careful verification of mathematical equations before posting.
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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.
 
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tmt said:
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
[math]\frac{2^1}{1!} + 1 = \frac{2}{1} + 1 = 2 + 1 = 3[/math]

[math]\frac{2^{1 + 1}}{1 + 1} = \frac{2^2}{2} = \frac{4}{2} = 2[/math]

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
 
tmt said:
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.
 
tmt said:
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

[math]\frac{2^2}{2!} + 1 = \frac{4}{2} + 1 = 2 + 1 = 3[/math]

[math]\frac{2^{2 + 1}}{(2 + 1)!} = \frac{2^3}{3!} = \frac{8}{6} = \frac{4}{3}[/math]

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

-Dan
 
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