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?

  • 4
    $\begingroup$ Yeah, it looks good. $\endgroup$ – Simply Beautiful Art Mar 28 '17 at 17:36
  • $\begingroup$ You're right. :) $\endgroup$ – pie314271 Mar 28 '17 at 17:37
  • $\begingroup$ Nice spot, looks like simply beautiful art to me $\endgroup$ – mrnovice Mar 28 '17 at 17:37
  • $\begingroup$ yes you are right, the result is $64$ $\endgroup$ – Dr. Sonnhard Graubner Mar 28 '17 at 17:38
  • $\begingroup$ Thanks a lot to all of you who have appreciated my efforts. $\endgroup$ – Saksham Mar 28 '17 at 18:59

Yes, this is correct.

Technically, I wouldn't call that intuition, I'd call it a proof.

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


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