First of all, is it called "convert" in English or there is a proper word for this purpose?
I am struggling with the algebra, please help me convert:
$(k+1)!\times(k+2)−1$ to $(k+2)!−1$
And what is the rule for this step?
Best wishes,
First of all, is it called "convert" in English or there is a proper word for this purpose?
I am struggling with the algebra, please help me convert:
$(k+1)!\times(k+2)−1$ to $(k+2)!−1$
And what is the rule for this step?
Best wishes,
$\rm \color{Blue}{(k+1)!}\cdot \color{Green}{(k+2)}=\color{Blue}{1\cdot2\cdot3\cdots k\cdot(k+1)}\cdot \color{Green}{(k+2)}=(k+2)!$
There's no special "rule" at play above; if you know the definition of the factorial this is immediate.
Without need, $k$ was assumed to be an integer here. One can instead use this definition $ z!=\Gamma(z+1)=\int_0^\infty t^{z} e^{-t}\, \mathrm{d}t. \! $ which converges absolutely, if the real part of the complex number z is positive $(\Re(z) > 0)$, to show that $ (k+2)!=\Gamma(k+3)=\int_0^\infty t^{k+2} e^{-t}\, \mathrm{d}t=\underbrace{\left[-t^{k+2}e^{-t}\right]_0^\infty}_{=0}+(k+2)\int_0^\infty t^{k+1} e^{-t}\, \mathrm{d}t =(k+2)\Gamma(k+2)=(k+2)\cdot (k+1)! $
For any positive integer $n$, $n!$ is defined as, $n!=n(n-1)\cdots1.$
From the above definition it follows that $n!=n(n-1)!,\forall n\in\mathbb{Z^+}.$
Hence it follows that $(k+1)!\times(k+2)−1=(k+2)!−1.$