For non-negative integers $s$ and $r$, let \[\binom{s}{r}= \begin{cases}\frac{s!}{r!(s-r)!} & \text{ if } r \leq s, \\ 0 & \text{ if } r>s. \end{cases}\] For positive integers $m$ and $n$, let \[g(m,n) = \sum_{p=0}^{m+n} \frac{f(m,n,p)}{\binom{n+p}{p}}\] where for any nonnegative integer $p$, \[f(m,n,p)=\sum_{i=0}^{p}\binom{m}{i}\binom{n+i}{p}\binom{p+n}{p-i}\] Then which of the following statements is/are TRUE?

(A) $g(m,n)=g(n,m)$ for all positive integers $m, n$; (B) $g(m,n+1)=g(m+1,n)$ for all positive integers $m, n$; (D) $g(2m,2n)=(g(m,n))^{2}$ for all positive integers $m, n$

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

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Binomial Theorem

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For non-negative integers

s

and

r

, let

(r s) = {\frac{s !}{r ! ( s - r )!} 0 if r \leq s, if r > s .

For positive integers

m

and

n

, let

g (m, n) = p = 0 \sum m + n \frac{f ( m , n , p )}{( p n + p )}

where for any nonnegative integer

p

f (m, n, p) = i = 0 \sum p (i m) (p n + i) (p - i p + n)

Then which of the following statements is/are TRUE?

(A) $g (m, n) = g (n, m)$ for all positive integers $m, n$ ✓ Correct
(B) $g (m, n + 1) = g (m + 1, n)$ for all positive integers $m, n$ ✓ Correct
(C) $g (2 m, 2 n) = 2 g (m, n)$ for all positive integers $m, n$
(D) $g (2 m, 2 n) = (g (m, n))^{2}$ for all positive integers $m, n$ ✓ Correct

Answer: (A), (B), (D)

Solution

Simplifying

f (m, n, p)

: each term

(i m) (p n + i) (p - i p + n)

can be rewritten as

(i m) (p n + p) (p - i n)

. Summing over

i

using Vandermonde's identity:

\sum_{i = 0}^{p} (i m) (p - i n) = (p m + n)

, so

f (m, n, p) = (p n + p) (p m + n)

. Thus

\frac{f ( m , n , p )}{( p n + p )} = (p m + n)

, and

g (m, n) = \sum_{p = 0}^{m + n} (p m + n) = 2^{m + n}

. Checking the options: (A)

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

✓, (B)

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

✓, (C)

g (2 m, 2 n) = 2^{2 m + 2 n} \neq = 2 \cdot 2^{m + n}

✗, (D)

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

✓.

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · One or more correct options

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JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Binomial Theorem

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