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

$$x=\sum _{0⩽i⩽j⩽2019}\left({}_j^{2019}\right)\left({}_i^j\right)$$

Commented by naka3546 last updated on 25/May/19

$$x=?$$

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

$$x=\sum _{j=0}^{2019}\left(\sum _{i=0}^j{C}_j^i\right)C_{2019}^j$$$$\sum _{i=0}^j{C}_j^i=\sum _{i=0}^j{C}_j^i{1}^i{1}^{j-i}=2^j\Rightarrow$$$$x=\sum _{j=0}^{2019}{C}_{2019}^j{2}^j{1}^{2019-j}$$$$=3^{2019}$$

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