1
$\begingroup$

(1)$(n+1)!−1+(n+1)×(n+1)!$

(2)$=(1+n+1)×(n+1)!−1$

(3)$=(n+2)×(n+1)!−1 $

(4)$=(n+2)!−1$

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

$\endgroup$
  • 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
3
$\begingroup$

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

$c = 1\times c$

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.