JEE Advance - Mathematics (2020 - Paper 2 Offline - No. 12)
For non-negative integers s and r, let
$$\left( {\matrix{ s \cr r \cr } } \right) = \left\{ {\matrix{ {{{s!} \over {r!(s - r)!}}} & {if\,r \le \,s,} \cr 0 & {if\,r\, > \,s} \cr } } \right.$$
For positive integers m and n, let
$$g(m,\,n) = \sum\limits_{p = 0}^{m + n} {{{f(m,n,p)} \over {\left( {\matrix{ {n + p} \cr p \cr } } \right)}}} $$
where for any non-negative integer p,
$$f(m,n,p) = \sum\limits_{i = 0}^p {\left( {\matrix{ m \cr i \cr } } \right)\left( {\matrix{ {n + i} \cr p \cr } } \right)\left( {\matrix{ {p + n} \cr {p - i} \cr } } \right)} $$
Then which of the following statements is/are TRUE?
$$\left( {\matrix{ s \cr r \cr } } \right) = \left\{ {\matrix{ {{{s!} \over {r!(s - r)!}}} & {if\,r \le \,s,} \cr 0 & {if\,r\, > \,s} \cr } } \right.$$
For positive integers m and n, let
$$g(m,\,n) = \sum\limits_{p = 0}^{m + n} {{{f(m,n,p)} \over {\left( {\matrix{ {n + p} \cr p \cr } } \right)}}} $$
where for any non-negative integer p,
$$f(m,n,p) = \sum\limits_{i = 0}^p {\left( {\matrix{ m \cr i \cr } } \right)\left( {\matrix{ {n + i} \cr p \cr } } \right)\left( {\matrix{ {p + n} \cr {p - i} \cr } } \right)} $$
Then which of the following statements is/are TRUE?
g(m, n) = g(n, m) for all positive integers m, n
g(m, n + 1) = g(m + 1, n) for all positive integers m, n
g(2m, 2n) = 2g(m, n) for all positive integers m, n
g(2m, 2n) = (g(m, n))2 for all positive integers m, n
Explanation
Since,
$$f(m,n,p) = \sum\limits_{i = 0}^p {\left( {\matrix{ m \cr i \cr } } \right)\left( {\matrix{ {n + i} \cr p \cr } } \right)\left( {\matrix{ {p + n} \cr {p - i} \cr } } \right)} $$
$$ = \sum\limits_{i = 0}^p {{{m!} \over {i!(m - i)!}} \times {{(n + i)!} \over {p!(n + i - p)!}} \times {{(p + n)!} \over {(p - i)!(n + i)!}}} $$
$$ = {{(n + p)!} \over {p!}}\sum\limits_{i = 0}^p {{}^m{C_i} \times {}^n{C_{p - i}}} $$
$$ = {}^{n + p}{C_p}{}^{m + n}{C_p}$$
Since, $$g(m,\,n) = \sum\limits_{p = 0}^{m + n} {{{f(m,n,p)} \over {\left( {\matrix{ {n + p} \cr p \cr } } \right)}}} $$
$$ = \sum\limits_{p = 0}^{m + n} {{{{}^{n + p}{C_p} \times {}^{m + n}{C_p}} \over {{}^{n + p}{C_p}}}} $$
$$ = \sum\limits_{p = 0}^{m + n} {{}^{m + n}{C_p}} = {2^{m + n}}$$
$$ \therefore $$ $$g(m,n) = g(n,m) = {2^{m + n}}$$,
and $$g(m,n + 1) = {2^{m + n + 1}} = g(m + 1,n)$$ and
$$g(2m,2n) = {2^{2(m + n)}} = {({2^{(m + n)}})^2} = {(g(m,n))^2}$$.
$$f(m,n,p) = \sum\limits_{i = 0}^p {\left( {\matrix{ m \cr i \cr } } \right)\left( {\matrix{ {n + i} \cr p \cr } } \right)\left( {\matrix{ {p + n} \cr {p - i} \cr } } \right)} $$
$$ = \sum\limits_{i = 0}^p {{{m!} \over {i!(m - i)!}} \times {{(n + i)!} \over {p!(n + i - p)!}} \times {{(p + n)!} \over {(p - i)!(n + i)!}}} $$
$$ = {{(n + p)!} \over {p!}}\sum\limits_{i = 0}^p {{}^m{C_i} \times {}^n{C_{p - i}}} $$
$$ = {}^{n + p}{C_p}{}^{m + n}{C_p}$$
Since, $$g(m,\,n) = \sum\limits_{p = 0}^{m + n} {{{f(m,n,p)} \over {\left( {\matrix{ {n + p} \cr p \cr } } \right)}}} $$
$$ = \sum\limits_{p = 0}^{m + n} {{{{}^{n + p}{C_p} \times {}^{m + n}{C_p}} \over {{}^{n + p}{C_p}}}} $$
$$ = \sum\limits_{p = 0}^{m + n} {{}^{m + n}{C_p}} = {2^{m + n}}$$
$$ \therefore $$ $$g(m,n) = g(n,m) = {2^{m + n}}$$,
and $$g(m,n + 1) = {2^{m + n + 1}} = g(m + 1,n)$$ and
$$g(2m,2n) = {2^{2(m + n)}} = {({2^{(m + n)}})^2} = {(g(m,n))^2}$$.
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