Hi, i'm having troubles understanding operations with faculty.
If:
(3k)! (k+1)! (2(k+1))!
__________________
k! (2k)! (3(k+1))!
Then why is this equal to:
(k+1)(2k+1)(2k+2)
________________
(3k+1)(3k+2)(3k+3)
Thanks for the help!
Hi, i'm having troubles understanding operations with faculty.
If:
(3k)! (k+1)! (2(k+1))!
__________________
k! (2k)! (3(k+1))!
Then why is this equal to:
(k+1)(2k+1)(2k+2)
________________
(3k+1)(3k+2)(3k+3)
Thanks for the help!
Hello, jones123!
I must assume that you know how to "cancel" factorials.
$\displaystyle \text{Show that: }\: \frac{\big(3k\big)!\,\big(k+1\big)!\,\big(2[k+1]\big)!}{k!\,\big(2k\big)!\,\big(3[k+1]\big)!} \;=\;\frac{(k+1)(2k+1)(2k+2)}{(3k+1)(3k+2)(3k+3)}$
$\displaystyle \text{We have: }\:\frac{(3k)!\,(k+1)!\,(2k+2)!}{k!\,(2k)!\,(3k+3) !} $
. . . . . $\displaystyle =\;\frac{(3k)!}{(3k+3)!}\cdot \frac{(k+1)!}{k!}\cdot\frac{(2k+2)!}{(2k)!}$
. . . . . $\displaystyle =\;\frac{1}{(3k+3)(3k+2)(3k+1)}\cdot\frac{k+1}{1} \cdot\frac{(2k+2)(2k+1)}{1} $
. . . . . $\displaystyle =\;\frac{(k+1)(2k+1)(2k+2)}{(3k+1)(3k+2)(3k+3)}$
But this reduces further . . .
. . . . . $\displaystyle =\;\frac{(k+1)\,(2k+1)\,2(k+1)}{(3k+1)\,(3k+2)\,3( k+1)}$
. . . . . $\displaystyle =\;\frac{2(k+1)(2k+1)}{3(3k+1)(3k+2)} $