Mathematics - Binomial Theorem Question with Solution | TestHub

$f\left(m,n,p\right)=\sum _{i=0}^p\left(\begin{array}{c}m\\ i\end{array}\right)\left(\begin{array}{c}n+i\\ p\end{array}\right)\left(\begin{array}{c}p+n\\ p-i\end{array}\right)$

$=\sum _{i=0}^p{}^{m}C_i·\frac{\left(n+i\right)!}{p!\left(n+i-p\right)!}·\frac{(p+n)!}{\left(p-i\right)!\left(n+i\right)!}$

$=\sum _{i=0}^p{}^{m}C_i·\frac{(p+n)!}{p!n!}·\frac{n!}{\left(n+i-p\right)!\left(p-i\right)!}$

$=\sum _{i=0}^p{}^{m}C_i.{}^{p+n}C_p·{}^{n}C_{p-i}$

${}^{p+n}C_p·\sum _{i=0}^p{}^{m}C_i·{}^{n}C_{p-i}$

$\left(\because \sum _{i=0}^p{}^{m}C_i·{}^{n}C_{p-i}={}^{m+n}C_p\right)$

$\therefore f\left(m,n,p\right)={}^{p+n}C_p.{}^{m+n}C_p$

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

$g\left(m,n\right)=\sum _{p=0}^{m+n}{}^{m+n}C_p=2^{m+n}$

(A) $g(2m, 2n)=2^{2m+2n}$

$2g\left(m,n\right)=2.2^{m+n}$

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

(C) $g\left(2m,2n\right)=2^{2m+2n}={\left(2^{m+n}\right)}^2={\left(g\left(m,n\right)\right)}^2$

(D) $g\left(m,n\right)=g\left(n,m\right)$.