Question

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Answer 1

\[\begin{align*}(x+h)^n&=\sum_{k=0}^n\binom nkx^{n-k}h^k\\ &=\binom n0x^nh^0+\binom n1x^{n-1}h^1+\cdots+\binom n{n-1}x^{n-(n-1)}h^{n-1}+\binom nnx^{n-n}h^n\\ &=x^n+nx^{n-1}h+\cdots+nxh^{n-1}+h^n \end{align*}\]

Answer 2

i read that and said to myself what grade or college do you go to and who made such a problem....

Answer 3

at the \[\left(\begin{matrix}n \\ n-1\end{matrix}\right)\]
wouldnt this be

\[\frac{ n! }{ (n-1)!(1) }\]

Answer 4

so how does that equal 1?

Answer 5

@zepdrix

Answer 6

At k=n-1 ?
\[\Large\rm \frac{n!}{(n-1)!}=\frac{n\cdot (n-1)!}{(n-1)!}\]It equals n, yes?

Answer 7

hmm

Answer 8

yes

Answer 9

If you look at some actual numbers it might make sense,\[\Large\rm 4!=4\cdot3\cdot2\cdot1=4\cdot3!\]
\[\Large\rm \frac{4!}{3!}=\frac{4\cdot3!}{3!}=\frac{4!(4-1)!}{3!}\]Understand how that works? :d
Peeling a factor out of the factorial.

Answer 10

but i thought the format was

\[\frac{ n! }{ k!(n-k)! }\]

Answer 11

\[\Large\rm \left(\begin{matrix}n \\ \color{orangered}{k}\end{matrix}\right)\quad=\quad\frac{ n! }{ \color{orangered}{k}!(n-\color{orangered}{k})! }\]
\[\Large\rm \left(\begin{matrix}n \\ \color{orangered}{n-1}\end{matrix}\right)\quad=\quad\frac{ n! }{ \color{orangered}{(n-1)}!(n-\color{orangered}{(n-1)})! }\]

Answer 12

@VPNetwork this is what i do on summer break

Answer 13

oh really? @TylerD

Answer 14

Ya looks like you've got the right formula :)\[\Large\rm \left(\begin{matrix}n \\ \color{orangered}{n-1}\end{matrix}\right)\quad=\quad\frac{ n! }{ \color{orangered}{(n-1)}!(n-\color{orangered}{n+1})! }\]

Answer 15

so that equals n?

Answer 16

ah i see now

Answer 17

if n! = 4 and plug it in, it will end up being 4 or (n). so ya it is n

Answer 18

so many variables i get lost but thanks

Answer 19

Try to remember how to relate this to Pascal's Triangle as well :)
Pascal's Triangle gives you the coefficients for binomial expansion.