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Question

asked Nov 4, 2023 174k views

1 Answer

Vladsiv · Answer 1 · 2023-11-09T05:22:08+0000

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

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

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

$\rm f(m,n,p) = \sum \limits_(i = 0)^(p) {}^(m) C_(i) \: \: {}^(n + i) C_(p) \: \: {}^(p + n) C_(p - i)$

$\rm {}^(m) C_(i) \: \: {}^(n + i) C_(p) \: \: {}^(p + n) C_(p - i)$

$\rm {}^(m) C_(i) ((n + i)!)/(p!(n - p + i)!) * ((n + p)!)/((p - i)!(n + i)!)$

$\rm {}^(m) C_(i) * ((n + p)!)/(p!) * (1!)/((n -p + i)!(p - i)!)$

$\rm {}^(m) C_(i) * ((n + p)!)/(p!n!) * (n!)/((n -p + i)!(p - i)!)$

$\rm {}^(m) C_(i) \: \: {}^(n + p) C_(p) \: \: {}^(n) C_(p - i) \: \: \{{}^(m) C_(i) \: \: {}^(n ) C_(p - i) = {}^(m + n) C_(p ) \}$

$\rm f(m,n,p) = {}^(n + p) C_(p){}^(m + n) C_(p)$

$\rm \frac{f(m,n,p)}{{}^(n + p) C_(p)} = {}^(m + n) C_(p)$

Now,

$\rm g(m,n) = \sum \limits_(p = 0)^(m + n) \frac{f(m,n,p)}{{}^(n + p) C_(p)}$

$\rm g(m,n) = \sum \limits_(p = 0)^(m + n) {}^(m + n) C_(p)$

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

$\rm(A) \: g(m,n) = q(n,m)$

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

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

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

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

$= \rm(g(m,n)) {}^(2)$