Question

for non-negative integers s,r let
$\sf \binom{s}{r} = \begin{cases} \sf \dfrac{s!}{r!(s-r)!} \:\;if\:\;\:r\leqslant s \\\\\sf 0 \:\;\:\:\;\:\:\:\;\:\:\;\:\:\:\;\:\:\;\:\: r>s \end{cases}$

for positive integers m and n, let
$\displaystyle\sf (m,n) \: \sum\limits_{p=0}^{m+n} \dfrac{f(m,n,p)}{\binom{n+p}{p}}$

where for any non negative integer p,

$\displaystyle\sf f(m,n,p) = \sum\limits_{i=0}^{p} \binom{m}{i} \binom{n+i}{p} \binom{p+n}{p-i}$

then which of the following statements is true?

(A) (m,n) = g(m,n) for all positive integers m,n

(B) (m,n+1) = g(m+i,n) for all positive integers m,n

(C) (2m,2n) = 2g(m,n) for all positive integers m,n

(D) (2m,2n) = (g(m,n))² for all positive integers m,n


→ appropriate answers needed.​

Answer 1

$\displaystyle\sf f(m,n,p) = \sum\limits_{i=0}^{p} {}^m C_i {}^{n+1} C_p \cdot {}^{p+n} C_{p-i}$

$\sf {}^m C_i \cdot {}^{n+1} C_p \cdot {}^{p+n} C_{p-1}$

$\sf = {}^m C_i \cdot \dfrac{(n+i)!}{p!(n-p+i)!} \times \dfrac{(n+p)!}{(p-i)!(n+i)!}$

$\sf = {}^m C_i \times \dfrac{n+p)!}{p!} \times \dfrac{1}{(n-p+i)!(p-i)!}$

$\sf = {}^m C_i \times \dfrac{(n+p)!}{p!n!} \times \dfrac{n!}{(n-p+i)!(p-i)!}$

$\sf = {}^m C_i \cdot {}^{n+p} C_p \cdot {}^n C_{p-1} \{ {}^m C_i \cdot {}^n C_{p-i} = {}^{m+n} C_p \}$

$\sf f(m,n,p) = {}^{n+p} C_p \cdot {}^{m+n} C_p$

$\sf \dfrac{f(m,n,p)}{{}^{n+p} C_p} = {}^{m+n} C_p$

Now

$\displaystyle\sf g(m,n) = \sum\limits_{p=0}^{m+n} \dfrac{f(m,n,p)}{{}^{n+p} C_p}$

$\displaystyle\sf g(m,n) = \sum\limits_{p=0}^{m+n} {}^{m+n} C_p$

$\displaystyle\sf g(m,n) = 2^{m+n}$

now check for options

$\sf (A) \:g(m,n) = g(n,m)$

$\sf (B) \: g(m,n+1) = 2^{m+n+1}$

$\sf g(m+n,n) = 2^{m+1+n}$

$\sf (D) \: g(2m,2n) = 2^{2m+2n}$

$\sf = (2^{m+n}){}^2$

$\sf = (g(m,n))^2$

$\rm \therefore$ Options A,B,D are correct.