For non-negative integers s and r,let binom{s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $f(m, n, p) = \sum_{i=0}^{p} \binom{m}{i} \binom{n+i}{p} \binom{p+n}{p-i}$.Using the identity $\binom{n+i}{p} \binom{p+n}{p-i} = \binom{n+p}

Vedclass · Accepted Answer

$A, B, D$

Vedclass · Answer

(A) Given $f(m, n, p) = \sum_{i=0}^{p} \binom{m}{i} \binom{n+i}{p} \binom{p+n}{p-i}$.Using the identity $\binom{n+i}{p} \binom{p+n}{p-i} = \binom{n+p}{p} \binom{n+i}{i}$,we have:$f(m, n, p) = \sum_{i=0}^{p} \binom{m}{i} \binom{n+p}{p} \binom{n+i}{i} = \binom{n+p}{p} \sum_{i=0}^{p} \binom{m}{i} \binom{n+i}{i}$.Using Vandermonde's Identity or combinatorial simplification,$\sum_{i=0}^{p} \binom{m}{i} \binom{n+i}{i} = \binom{m+n+1}{p}$ is not directly applicable,but rather $\sum_{i=0}^{p} \binom{m}{i} \binom{n+i}{p-i}$ simplifies to $\binom{m+n+p}{p}$.Actually,$f(m, n, p) = \binom{n+p}{p} \binom{m+n}{p}$.Thus,$\frac{f(m, n, p)}{\binom{n+p}{p}} = \binom{m+n}{p}$.Then $g(m, n) = \sum_{p=0}^{m+n} \binom{m+n}{p} = 2^{m+n}$.Checking the options:$(A)$ $g(m, n) = 2^{m+n}$ and $g(n, m) = 2^{n+m}$,so $g(m, n) = g(n, m)$ is $TRUE$.$(B)$ $g(m, n+1) = 2^{m+n+1}$ and $g(m+1, n) = 2^{m+1+n}$,so $g(m, n+1) = g(m+1, n)$ is $TRUE$.$(C)$ $g(2m, 2n) = 2^{2m+2n}$ and $2g(m, n) = 2 \cdot 2^{m+n} = 2^{m+n+1}$,so $g(2m, 2n) 
eq 2g(m, n)$.$(D)$ $g(2m, 2n) = 2^{2m+2n} = (2^{m+n})^2 = (g(m, n))^2$,so $g(2m, 2n) = (g(m, n))^2$ is $TRUE$.Therefore,the correct statements are $A, B, D$.

Similar Questions

Let the coefficient of $x^{r}$ in the expansion of $(x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3}(x+2)^2+\ldots+(x+2)^{n-1}$ be $\alpha_{r}$. If $\sum_{r=0}^{n-1} \alpha_{r}=\beta^{n}-\gamma^{n}$,where $\beta, \gamma \in N$,then the value of $\beta^2+\gamma^2$ equals:

Let $(1+x+x^2)^9=a_0+a_1 x+a_2 x^2 +\ldots+a_{18} x^{18}$. Then

Let $M = 2^{30} - 2^{15} + 1$. When $M^2$ is expressed in base $2$,the number of $1$'s in its binary representation is:

Let $R=(5 \sqrt{5}+11)^{2 n+1}$ and $f=R-[R]$,where $[x]$ denotes the greatest integer less than or equal to $x$,then $R f=$

Suppose $2-p, p, 2-\alpha, \alpha$ are the coefficients of four consecutive terms in the expansion of $(1+x)^n$. Then the value of $p^2-\alpha^2+6\alpha+2p$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module