Operations with factorials

Dec 2012
19
0
be
Hi, i'm having troubles understanding operations with faculty.

If:

(3k)! (k+1)! (2(k+1))!
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k! (2k)! (3(k+1))!

Then why is this equal to:

(k+1)(2k+1)(2k+2)
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(3k+1)(3k+2)(3k+3)

Thanks for the help!
 
Jun 2012
45
17
France
Note that
\(\displaystyle (3(k+1))!=(3k+3)!=(3k+3)(3k+2)(3k+1)(3k)!\).
And do it again with 2 instead of 3.
 
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Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
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 . . .

. . . . . \(\displaystyle =\;\frac{(k+1)\,(2k+1)\,2(k+1)}{(3k+1)\,(3k+2)\,3(k+1)}\)

. . . . . \(\displaystyle =\;\frac{2(k+1)(2k+1)}{3(3k+1)(3k+2)} \)
 
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