# Is this result based on factorials true?

While solving some questions based on factorials, I noticed a pattern and sat down to prove what I had been observing, for all $n$.

I solved some questions like this one:

Solve for $x$ - $$\frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!}$$

On computing, I found out that the value of $x$ came out to be $8^2 = 64$.

On solving some other similar questions, I noticed that the value of $x$ came out to be the square of the denominator of $x$ itself. In other words, I can express it as follows -

Solve for $x$ - $$\frac{1}{n!} + \frac{1}{(n+1)!} = \frac{x}{(n+2)!}$$

So, the value of $x$ in such situations came out to $(n+2)^2$. Here is how I proved it -

$(n+1)! = (n+1).n!$
$(n+2)! = (n+2).(n+1).n!$

So, LHS becomes, $\frac{n+2}{(n+1).n!} = \frac{x}{(n+2).(n+1).n!}$

$\implies x = (n+2)^2$

Is my intuition correct? The pattern which I noticed led me to this result. Is this result valid?

• Yeah, it looks good. – Simply Beautiful Art Mar 28 '17 at 17:36
• You're right. :) – pie314271 Mar 28 '17 at 17:37
• Nice spot, looks like simply beautiful art to me – mrnovice Mar 28 '17 at 17:37
• yes you are right, the result is $64$ – Dr. Sonnhard Graubner Mar 28 '17 at 17:38
• Thanks a lot to all of you who have appreciated my efforts. – Saksham Mar 28 '17 at 18:59

Essentially, you are seeing $m(m-1) + m=m^2$, where $m=n+2$.