(3)$=(n+2)×(n+1)!−1 $


I understand how step 4 derived from 3, but I am confused on how does step 2 derived from step 1? thank you

  • 3
    $\begingroup$ Factor out the $(n+1)!$ from the two terms that have it. So, use commutativity of addition and subtraction to move the $-1$ to the end, then you have $1\times (n+1)!+(n+1)\times (n+1)!-1$. $\endgroup$ – InterstellarProbe May 2 at 20:06
  • $\begingroup$ Take out common factor $(n+1)!$ $\endgroup$ – lab bhattacharjee May 2 at 20:07

$a\times c + b\times c = (a+b)\times c$

$c = 1\times c$

Now, use the above with $1$ in place of $a$, with $(n+1)$ in place of $b$ and $(n+1)!$ in place of $c$ to get:

$(n+1)! + (n+1)\times (n+1)! = 1\times (n+1)! + (n+1)\times (n+1)! = (1+n+1)\times (n+1)!$


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