Question

JEE ADVANCED MATHS QUESTION SEPTEMBER 2020

Answer 1

For non negative integers S and r, let

$\binom{s}{r}$ = s!/r!(s - r)! if r ≤ s

= 0 if r > s

for positive integers m and n let,

$(m,n)\Sigma_{p=0}^{m+n}\frac{f(m,n,p)}{\binom{n+p}{p}$

where for any non negative integers p,

$f(m,n,p)=\Sigma_{i=0}^p\binom{m}{i}\binom{n+i}{p}\binom{p+n}{p-i}$

Then which of the following statements is/are TRUE ?

solution : f(m, n, p) = $\Sigma_{i=0}^p$$^mC_i$$^{n+i}C_p$$^{p+n}C_{p-i}$

= $^mC_i$ (n + i)!/{p! (n + i - p)! × (n + p)!/(p - i)! (n + p - p + i)!

= $^mC_i$ × (n + i)/p! (n - p + i)! × (n + p)!/(p - i)! (n + i)!

= $^mC_i$ × (n + p)!/p! × 1/(n - p + i)! (p - i)!

= $^mC_i$ × (n + p)!/(p! n !) × n!/(n - p + i)!(p - i)!

= $^mC_i$$^{n+p}C_p$$^nC_{p-i}$

here $^mC_i$ . $^nC_{p-i}$ = $^{m+n}C_p$

so, f(m, n, p) = $^{n+p}C_p$$^{m+n}C_p$

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

Now, g(m, n) = $\Sigma_{p=0}^{m+n}\frac{f(m,n,p)}{^{n+p}C_p}=\Sigma_{p=0}^{m+n}$$^{m+n}C_p$

so, g(m, n) = 2^(m + n)

now let's check all options.

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

Therefore option (A) is correct choice.

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

g(m + 1, n) = 2^(m + n + 1) hence option (B) is also correct.

(2m, 2n) = 2^(2m + 2m) = 2² 2^(m + n) = 4g(m, n) , option (C) is incorrect

(2m, 2n) = 2^(2m + 2m) = [2^(m + n)]² = g(m, n)² , option (D) is correct.

Therefore options (A), (B), (D) are correct choices.