help with factorials

[lost-and-found]

can someone tell me how [(n+1)!]^2 = ((n+1)^2)(n!)^2 ? I'm in that part of calculus where we do the stupid sequences and series, and I see a lot of factorials, and i have no idea how to manipulate them properly

[shawshank62]

can someone tell me how [(n+1)!]^2 = ((n+1)^2)(n!)^2 ? I'm in that part of calculus where we do the stupid sequences and series, and I see a lot of factorials, and i have no idea how to manipulate them properly

great, now i have something to look forward to this semester in calculus

[lost-and-found]

oh, I promise you, it's all FUN, FUN, FUN

[Optimus Prime]

Jesus, why do you need that stuff in life?

[lost-and-found]

you don't, but they force you to learn it anyways

[jkrohn]

Start with the Right Side, make it equal the left.

Right Side.
n! = 1*2*3*......n
(n!)^2 = (1*2*3....n)(1*2*3....n)
(n+1)^2 = (n+1)(n+1)

=> (n+1)(n+1)(1*2*3....n)(1*2*3...n)
=> (n+1)(1*2*3....n)(n+1)(1*2*3....n)
=> (1*2*3.....n)(n+1) * (1*2*3.....n)(n+1)
=> (1*2*3.....n*n+1) * (1*2*3....n*n+1)
=> (1*2*3.....n*n+1)^2
=> ((n+1)!)^2

Follow that?
Jkrohn

[tmx468]

[shudders]... I remember them bad old days.

L.A.F, you have my deepest sympathies!

Nice workthrough though jkrohn!

[Theophylact]

(n+1)! = (n!)(n+1) by definition. (Think about it: the product of the first n positive integers, times the (n+1)st integer, is the product of the first n+1 integers.)

Now square both sides. Presto!

[lost-and-found]

yes, I'm kind off ashamed for asking this question now, I looked at the factorial a bit more closer and then I realized it was just a simjple factoring out of (n+1)^2....thanks for the help though...

BTW, I still ahte series, sequences and factorials

[Prison Kids]

Ouch! time for a nap...