Tuesday, 27 October 2015

Monday, 26 October 2015

Grove HSC 3U book EX 10.1 Q15 Pascal Triangle identity

Prove $\left({n\atop k}\right)=\left({n-1\atop k-1}\right)+\left({n-1\atop k}\right)$


LHS $=\left({n\atop k}\right)={n!\over k! (n-k)!}$

RHS $=\left( {n-1\atop k-1}\right)+\left({n-1 \atop k}\right)$

$={(n-1)!\over (k-1)! (n-1-(k-1))!}+ {(n-1)!\over k! (n-1-k)!}$

$={(n-1)!\over (k-1)! (n-k)!}+ {(n-1)!\over k! (n-k-1)!}$

$={k(n-1)!\over k(k-1)! (n-k)!}+ {(n-1)!(n-k)\over k! (n-k)(n-k-1)!}$

$={k(n-1)!\over k! (n-k)!}+ {(n-1)!(n-k)\over k! (n-k)!}$

$={k(n-1)! + (n-1)!(n-k)\over k! (n-k)!}$

$={(n-1)!(k + (n-k))\over k! (n-k)!}$

$={(n-1)!n\over k! (n-k)!}$

$={n!\over k! (n-k)!}$
=LHS

as required.

FACTORISE $K! + (K+1)! $

FACTORISE 

$K! + (K+1)!$



$=  K!  +   (K+1)K!$

$= K!  (1 + K + 1)$

$= K! (K+2)$