x-0-i-j-2019-j-2019-i-j-

Question Number 60725 by naka3546 last updated on 25/May/19

x = \sum_{0 ⩽ i ⩽ j ⩽ 2019} (_{j}^{2019}) (_{i}^{j})

Commented by naka3546 last updated on 25/May/19

x = ?

Commented by Mr X pcx last updated on 25/May/19

x = \sum_{j = 0}^{2019} (\sum_{i = 0}^{j} C_{j}^{i}) C_{2019}^{j}

\sum_{i = 0}^{j} C_{j}^{i} = \sum_{i = 0}^{j} C_{j}^{i} 1^{i} 1^{j - i} = 2^{j} \Rightarrow

x = \sum_{j = 0}^{2019} C_{2019}^{j} 2^{j} 1^{2019 - j}

= 3^{2019}