J jones123 Dec 2012 19 0 be Jun 5, 2013 #1 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!

K Kmath Jun 2012 45 17 France Jun 5, 2013 #2 Note that \(\displaystyle (3(k+1))!=(3k+3)!=(3k+3)(3k+2)(3k+1)(3k)!\). And do it again with 2 instead of 3. Reactions: 1 person

Note that \(\displaystyle (3(k+1))!=(3k+3)!=(3k+3)(3k+2)(3k+1)(3k)!\). And do it again with 2 instead of 3.

S Soroban MHF Hall of Honor May 2006 12,028 6,341 Lexington, MA (USA) Jun 5, 2013 #3 Hello, jones123! I must assume that you know how to "cancel" factorials. \(\displaystyle \text{Show that: }\: \frac{\big(3k\big)!\,\big(k+1\big)!\,\big(2[k+1]\big)!}{k!\,\big(2k\big)!\,\big(3[k+1]\big)!} \;=\;\frac{(k+1)(2k+1)(2k+2)}{(3k+1)(3k+2)(3k+3)}\) Click to expand... \(\displaystyle \text{We have: }\:\frac{(3k)!\,(k+1)!\,(2k+2)!}{k!\,(2k)!\,(3k+3)!} \) . . . . . \(\displaystyle =\;\frac{(3k)!}{(3k+3)!}\cdot \frac{(k+1)!}{k!}\cdot\frac{(2k+2)!}{(2k)!}\) . . . . . \(\displaystyle =\;\frac{1}{(3k+3)(3k+2)(3k+1)}\cdot\frac{k+1}{1} \cdot\frac{(2k+2)(2k+1)}{1} \) . . . . . \(\displaystyle =\;\frac{(k+1)(2k+1)(2k+2)}{(3k+1)(3k+2)(3k+3)}\) But this reduces further . . . . . . . . \(\displaystyle =\;\frac{(k+1)\,(2k+1)\,2(k+1)}{(3k+1)\,(3k+2)\,3(k+1)}\) . . . . . \(\displaystyle =\;\frac{2(k+1)(2k+1)}{3(3k+1)(3k+2)} \) Reactions: 1 person

Hello, jones123! I must assume that you know how to "cancel" factorials. \(\displaystyle \text{Show that: }\: \frac{\big(3k\big)!\,\big(k+1\big)!\,\big(2[k+1]\big)!}{k!\,\big(2k\big)!\,\big(3[k+1]\big)!} \;=\;\frac{(k+1)(2k+1)(2k+2)}{(3k+1)(3k+2)(3k+3)}\) Click to expand... \(\displaystyle \text{We have: }\:\frac{(3k)!\,(k+1)!\,(2k+2)!}{k!\,(2k)!\,(3k+3)!} \) . . . . . \(\displaystyle =\;\frac{(3k)!}{(3k+3)!}\cdot \frac{(k+1)!}{k!}\cdot\frac{(2k+2)!}{(2k)!}\) . . . . . \(\displaystyle =\;\frac{1}{(3k+3)(3k+2)(3k+1)}\cdot\frac{k+1}{1} \cdot\frac{(2k+2)(2k+1)}{1} \) . . . . . \(\displaystyle =\;\frac{(k+1)(2k+1)(2k+2)}{(3k+1)(3k+2)(3k+3)}\) But this reduces further . . . . . . . . \(\displaystyle =\;\frac{(k+1)\,(2k+1)\,2(k+1)}{(3k+1)\,(3k+2)\,3(k+1)}\) . . . . . \(\displaystyle =\;\frac{2(k+1)(2k+1)}{3(3k+1)(3k+2)} \)