1. ## Binomial Question

Show that for all positiver integers n,

x[ (1+x)^n-1 + (1+x)^n-2 +...+ (1+x)^2 + (1+x) + 1 ] = (1+x)^n-1

ii) Hence show that for 1< and= k < and= n,

(n-1)C(k-1) + (n-2)C(k-2) + (n-3)C(k-3) + ... + (k-1)C(k-1) = nCk

is nCr (the combinations thing)

I can't do the second part. Help pls

2. Originally Posted by noobonastick
Show that for all positiver integers n,

x[ (1+x)^n-1 + (1+x)^n-2 +...+ (1+x)^2 + (1+x) + 1 ] = (1+x)^n-1

ii) Hence show that for 1< and= k < and= n,

(n-1)C(k-1) + (n-2)C(k-2) + (n-3)C(k-3) + ... + (k-1)C(k-1) = nCk

is nCr (the combinations thing)

I can't do the second part. Help pls
HI

For part (1) , Its a geometric progression with first term , $a=(1+x)^{n-1}$ , $r=(1+x)^{-1}$

So using the summation formula ,

$\frac{(1+x)^{n-1}[1-(1+x)^{-n}]}{1-(1+x)^{-1}}$

$\frac{(1+x)^{n-1}[1-(1+x)^{-n}]}{x(1+x)^{-1}}$

$
\frac{(1+x)^n\cdot (1+x)^{-1}[\frac{(1+x)^n-1}{(1+x)^n}]}{x(1+x)^{-1}}
$

Do some cancellings , then you are left with

$\frac{(1+x)^n-1}{x}$

Both the x's (there is one more x above which i din include in my working) would cancel off .

Sorry , just realised you are asking help on the second part .. i am gonna post this anyways .

3. Originally Posted by noobonastick
Show that for all positiver integers n,

x[ (1+x)^n-1 + (1+x)^n-2 +...+ (1+x)^2 + (1+x) + 1 ] = (1+x)^n-1

ii) Hence show that for 1< and= k < and= n,

(n-1)C(k-1) + (n-2)C(k-2) + (n-3)C(k-3) + ... + (k-1)C(k-1) = nCk

is nCr (the combinations thing)

I can't do the second part. Help pls
For the second part, find the coefficient of $x^k$ on the LHS and RHS of the identity in the first part.