value of $f(2019)$ in binomial expression

Question

if $\displaystyle f(n)=\mathop{\sum}_{i>j\geq 0}\binom{n+1}{i}\binom{n}{j}.$ Then value of $f(2019)$ is

what i try $\displaystyle f(n)=\binom{n+1}{1}\binom{n}{0}+\binom{n+1}{2}\bigg(\binom{n}{0}+\binom{n}{1}\bigg)+\cdots + \binom{n+1}{n+1}\bigg(\binom{n}{0}+\binom{n}{1}+\cdots +\binom{n}{n}\bigg)$

How do i simplify it help me please

Answer 1

$$ \begin{align} \sum_{i\gt j}\binom{n+1}{i}\binom{n}{j} &=\sum_{i\gt j}\left[\binom{n}{i}+\binom{n}{i-1}\right]\binom{n}{j}\tag1\\ &=\sum_{i\gt j}\binom{n}{i}\binom{n}{j}+\sum_{i\ge j}\binom{n}{i}\binom{n}{j}\tag2\\ &=\sum_{i,j}\binom{n}{i}\binom{n}{j}\tag3\\[6pt] &=2^n\cdot2^n\tag4\\[15pt] &=4^n\tag5 \end{align} $$ Explanation:
$(1)$: Pascal's Rule
$(2)$: substitute $i\mapsto i+1$ in the right-hand sum
$(3)$: swap $i$ and $j$ by symmetry in the right-hand sum
$(4)$: $\sum\binom{n}{i}=(1+1)^n=2^n$
$(5)$: simplify

Answer 2

Hint: First show that $\sum_{j=0}^n\binom{n}{j}=2^n$.

Answer 3

Let's first calculate $2f(n)$:

$$ \begin{align} 2f(n) &= \sum_{j=0}^{n}\sum_{i=j+1}^{n+1}\binom{n+1}{i}\binom{n}{j} + \sum_{j=0}^{n}\sum_{i=j+1}^{n+1}\binom{n+1}{i}\binom{n}{n-j}\\ &= \sum_{j=0}^{n}\binom{n}{j}\sum_{i=j+1}^{n+1}\binom{n+1}{i} + \sum_{j=0}^{n}\binom{n}{n-j}\sum_{i=j+1}^{n+1}\binom{n+1}{i}\\ &= \sum_{j=0}^{n}\binom{n}{j}\sum_{i=j+1}^{n+1}\binom{n+1}{i} + \sum_{j=0}^{n}\binom{n}{j}\sum_{i=n-j+1}^{n+1}\binom{n+1}{i}\\ &= \sum_{j=0}^{n}\binom{n}{j}\sum_{i=j+1}^{n+1}\binom{n+1}{i} + \sum_{j=0}^{n}\binom{n}{j}\sum_{i=0}^{j}\binom{n+1}{i}\\ &= \sum_{j=0}^{n}\binom{n}{j}\sum_{i=0}^{n+1}\binom{n+1}{i}\\ &= 2^n2^{n+1} = 2^{2n+1} \end{align} $$

Therefore: $f(n) = 2^{2n} = 4^n$ and thus $f(2019) = 4^{2019}$.