Multiply Factorial
I have a small comprehension gap with an easy equation. I have following term and I don’t know how to multiply it correctly. $ (n+1)(n+1)!+(n+1)!-1 $
One intermediate step must be. $ (n+2)(n+1)!-1 $
The result should be $ (n+2)!-1 $.
How do I multiply the term correct? Is the attempt correct to multiply binomial series to $ (n+1)^2n!+(n+1)!-1 $?
It would awesome, if someone could help me.
$\endgroup$ 02 Answers
$\begingroup$$$(n+1)\color{red}{(n+1)!}+\color{red}{(n+1)!}-1$$ $$=(n+1)![(n+1)+1]-1$$ $$=(n+2)(n+1)!-1$$ $$=(n+2)!-1$$
$\endgroup$ 6 $\begingroup$You would have no trouble with $(5)(n+1)!+(7)(n+1)!=(12)(n+1)!$ ($5$ apples plus $7$ apples equals $12$ apples.)
Maybe you would have a little trouble with $(5)(n+1)!+(n+1)!$, but not if you rewrite it as $(5)(n+1)! +(1)(n+1)!$ ($5$ apples plus $1$ apple equals $6$ apples).
Now let's look at $(n+1)(n+1)! +(n+1)!$. Rewrite this as $(n+1)(n+1)! +(1)(n+1)!$. We have $n+1$ apples plus $1$ apple is $n+2$ apples, so the sum is $(n+2)(n+1)!$, which can be rewritten as $(n+2)!$.
$\endgroup$ 1
