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Question Number 186630 by DAVONG last updated on 07/Feb/23

Answered by mr W last updated on 07/Feb/23

$$a_k=1+2+3+...+k=\frac{k(k+1)}{2}=\frac{k^2+k}{2}$$$$\underset{k=1}{\sum ^n}{a}_k=\frac{1}{2}(\underset{k=1}{\sum ^n}{k}^2+\underset{k=1}{\sum ^n}k)$$$$=\frac{n(n+1)(2n+1)}{12}+\frac{n(n+1)}{4}$$$$=\frac{n(n+1)(n+2)}{6}$$$$withn=2022:$$$$\frac{n(n+1)(n+2)}{6}$$$$=\frac{2022\times 2023\times 2024}{6}=1379864024$$

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