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Question Number 94690 by abony1303 last updated on 20/May/20

Commented by abony1303 last updated on 20/May/20

$$PlsHelp!$$

Commented by prakash jain last updated on 20/May/20

$$$$$$\mathrm{First}\mathrm{term}{a}_1=1$$$$a_2,a_3=2$$$$a_4,a_5,a_6=3$$$$a_k=n\mathrm{for}\frac{n(n-1)}{2}<k⩽\frac{n(n+1)}{2}$$$$k=2019$$$$n=64$$$$\frac{64\times 63}{2}=2016<2019⩽\frac{64\times 65}{2}=2080$$

Commented by abony1303 last updated on 20/May/20

$$wherecanIlearnthiskindofquestions$$$$Imean{what}^{\prime }sthenameofthistheme?$$

Commented by mr W last updated on 21/May/20

$$a_k=\lceil \frac{\sqrt{8k+1}-1}{2}\rceil$$$$a_{2019}=\lceil \frac{\sqrt{8\times 2019+1}-1}{2}\rceil =64$$

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