Binomial question (1 Viewer)

Sy123 · Oct 25, 2015

Answered here: http://community.boredofstudies.org...1695/hsc-2015-3u-marathon-48.html#post7057100

------------------------------------

$\\ $i) Using the definition$ \ \binom{n}{r} := \frac{n!}{(n-r)!r!} \\ $So,$ \ \binom{n}{r} = \frac{n!}{(n-r)! r!} = \frac{n!}{(n- (n-r))! (n-r)!} = \binom{n}{n-r}$

$\\ $ii) Transform the expression$ \ f(r) \ $into an easier to deal form, that is, replace every$ \ \binom{n}{r+k} \ $with$ \ \binom{n}{n - (r+k)}$

$\\ $So,$ \ f(r) = \binom{n}{0} \binom{n}{n-r} + \binom{n}{1} \binom{n}{n - r - 1} + \binom{n}{2} \binom{n -r - 2} + \dots + \binom{n}{n-r} \binom{0}$

$\\ $Consider the co-efficient of$ \ x^{n-r} \ $on both sides of the equation$ \ (1+x)^{2n} = (1+x)^n (1+x)^n \\\\ $On the left-hand-side the co-efficient is$ \ \binom{2n}{n-r}$

$\\ $On the right-hand-side, the co-efficient is arrived at by taking the coefficient of$ \ x^k \ $from the first$ \ (1+x)^n \ $and$ \ x^{n-r-k} \ $from the second$ \ (1+x)^n \ $for all$ \ k = 0,1,2, \dots, n-r$

$\\ $So,$ \ \binom{2n}{n-r} = \sum_{k=0}^{n-r} \underbrace{\binom{n}{k} \binom{n}{n-r-k}}_{\text{Take} \ k \ \text{from the first bracket, and} \ n-r-k \ \text{from the second}}$

$\\ f(r) = \sum_{k=0}^{n-r}$

$\Rightarrow \ f(r) = \binom{2n}{n-r}$

$\\ $iii) Consider the coefficient of$ \ x^n \ $in the expression$ \ (1+x)^{3n} = (1+x)^n (1+x)^{2n}$

$\\ $Take$ \ x^k \ $from the first bracket, and$ \ x^{n-k} \ $from the second one, meaning$ \ \binom{3n}{n} = \sum_{k=0}^n \binom{n}{k} \binom{2n}{n-k} = \sum_{k=0}^n \binom{n}{k} f(k)$

Binomial question (1 Viewer)

duxxx

New Member

Sy123

This too shall pass

Users Who Are Viewing This Thread (Users: 0, Guests: 1)