Binomial Theorem 2

[IloveManUtd]

The expansion of (1+kx)[sup:2mpidhus]n[/sup:2mpidhus], where n is a positive integer, is 1 + 8x + 120k[sup:2mpidhus]2[/sup:2mpidhus]x[sup:2mpidhus]2[/sup:2mpidhus] + hx[sup:2mpidhus]3[/sup:2mpidhus] +......
Calculate the values of n, h and k

I've already done: (1+kx)[sup:2mpidhus]n[/sup:2mpidhus] = 1[sup:2mpidhus]n[/sup:2mpidhus] + [[sup:2mpidhus]n[/sup:2mpidhus]C[sub:2mpidhus]1[/sub:2mpidhus](1[sup:2mpidhus]n-1[/sup:2mpidhus])(kx)] + [[sup:2mpidhus]n[/sup:2mpidhus]C[sub:2mpidhus]2[/sub:2mpidhus](1[sup:2mpidhus]n-2[/sup:2mpidhus])(kx)[sup:2mpidhus]2[/sup:2mpidhus]] + [[sup:2mpidhus]n[/sup:2mpidhus]C[sub:2mpidhus]3[/sub:2mpidhus](1[sup:2mpidhus]n-3[/sup:2mpidhus])(kx])[sup:2mpidhus]3[/sup:2mpidhus]].........

How do I convert the [sup:2mpidhus]n[/sup:2mpidhus]C[sub:2mpidhus]1[/sub:2mpidhus], [sup:2mpidhus]n[/sup:2mpidhus]C[sub:2mpidhus]2[/sub:2mpidhus].... into something workable, THX.

 

[soroban]

Hello, IloveManUtd!

\(\displaystyle \text{The expansion of }(1+kx)^n\text{, where }n\text{ is a positive integer, is: }\;1 + 8x + 120k^2x^2 + hx^3 + \hdots\)

\(\displaystyle \text{Calculate the values of }n,\,h,\text{ and }k.\)

\(\displaystyle \text{I've already done: }\;(1+ kx)^n \:=\: 1 + (^nC_1)(kx) + (^nC_2)(kx)^2 + (^nC_3)(kx)^3 + \hdots\) . Good!

\(\displaystyle \text{How do I convert the }^nC_1,\;^nC_2,\:\hdots \text{ into something workable?}\)

\(\displaystyle \text{We're expected to know that: }\;^nC_r \:=\:\frac{n!}{r!(n-r)!}\)


\(\displaystyle \text{Then we have: }\;\begin{Bmatrix} ^nC_1 & = & n \\ ^nC_2 &=& \dfrac{n(n-1)}{2} \\ \\[-2mm] ^nC_3 &=& \dfrac{n(n-1)(n-2)}{6} \end{Bmatrix}\)


\(\displaystyle \text{Can you finish it?}\)

 

[BigGlenntheHeavy]

\(\displaystyle soroban, \ good \ show \ on \ this \ one.\)

\(\displaystyle I, \ myself \ was \ originally \ stumped \ on \ this \ one \ until \ you \ interjected \ combinatorial\)

\(\displaystyle notation, \ then \ it \ all \ came \ back, \ again \ good \ show.\)

 

[IloveManUtd]

Thanks man, soroban