Binomial Expansion Help (1 Viewer)

[micuzzo]

can someone please explain why (7-k+1)! = (7-k+1)(7-k)(7-k-1)! if the one has already been removed from the first term (7-k+1) to get (7-k) but then 1 is still removed from (7-k) to get (7-k-1) shoudnt it be (7-k-2)??? or am i reading it wrong.

 

[tommykins]

let 7-k = a
(a+1)! = (a+1)a! = (a+1)(a)(a-1)!

replace the a's with 7-k and that's what you get.

 

[micuzzo]

why does (a+1)!=(a+1)a!... sorry im a lil confused

 

[tommykins]

it's the rule of factorials.

n! = 1*2*3*4...*n
so (n+1)! = 1*2*3*4...*n*(n+1) = n!(n+1)

 

[micuzzo]

ahh yeah... i think im confusin my self with n!=n(n-1)(n-2)(n-2)...

binomials... its been a while

 

[addikaye03]

why does (a+1)!=(a+1)a!... sorry im a lil confused

(a+1) is the term above a, therefore (a+1)!=(a+1)(a)(a-1)...3.2.1, similarly this can be expressed as (a+1)a! ( since a!=multipllication of all terms below a)

 

[Trebla]

Using a simple numerical example:
If
6! = 1 x 2 x 3 x 4 x 5 x 6
and
5! = 1 x 2 x 3 x 4 x 5
then,
6! = (1 x 2 x 3 x 4 x 5) x 6
= 5! x 6
Similarly, if
4! = 1 x 2 x 3 x 4
then,
6! = (1 x 2 x 3 x 4) x 5 x 6
= 4! x 5 x 6

 

[Valupatitta]

i really cant do proving binomials for shit. i did a mock test today and i only got 73%. i was bawling my eyes out :'(

 

[addikaye03]

Using a simple numerical example:
If
6! = 1 x 2 x 3 x 4 x 5 x 6
and
5! = 1 x 2 x 3 x 4 x 5
then,
6! = (1 x 2 x 3 x 4 x 5) x 6
= 5! x 6
Similarly, if
4! = 1 x 2 x 3 x 4
then,
6! = (1 x 2 x 3 x 4) x 5 x 6
= 4! x 5 x 6

haha that was better explaination than mine. Same concept tho