Binomial Question (1 Viewer)

[Andy005]

Use (1+x)^n=(1+x)(1+x)^n-1 to prove that nCk= n-1Ck + n-1Ck-1

Thanks.

 

[pikachu975]

Try expand and equate coefficients of x^k on both sides (which is seen by the nCk on the LHS.

I'll write a solution later if needed

 

[Andy005]

Try expand and equate coefficients of x^k on both sides (which is seen by the nCk on the LHS.

I'll write a solution later if needed

I understand what you mean, but i'm unsure on how to get the coeff of x^k on the RHS.

Thanks.

 

[pikachu975]

I understand what you mean, but i'm unsure on how to get the coeff of x^k on the RHS.

Thanks.

You can see you have a (1+x) so for the (1+x)^(n-1) you need an x^(k-1) and an x^k because if you expand (1+x)(1+x)^(n-1) you can see the 1 will multiply with the term with x^k and the x will multiply with the term x^(k-1) to give x^k.

 

[Andy005]

You can see you have a (1+x) so for the (1+x)^(n-1) you need an x^(k-1) and an x^k because if you expand (1+x)(1+x)^(n-1) you can see the 1 will multiply with the term with x^k and the x will multiply with the term x^(k-1) to give x^k.

So if i were to expand (1+x)^n-1 would it be,

n-1C0 + n-1C1x + n-1C2x^2 + ... + n-1Ck-1x^k-1 + n-1Ckx^k + ... + n-1Cn-1x^n-1?

 

[pikachu975]

So if i were to expand (1+x)^n-1 would it be,

n-1C0 + n-1C1x + n-1C2x^2 + ... + n-1Ck-1x^k-1 + n-1Ckx^k + ... + n-1Cn-1x^n-1?

Yep that's right

 

[Andy005]

Yep that's right

Thanks for your help.