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Question Number 93091 by Maclaurin Stickker last updated on 10/May/20

$$Findthesum:$$$$\underset{n=1}{\sum ^{2020}}{2}^{2^n}$$

Commented by MJS last updated on 11/May/20

$$2^{2^{2020}}\approx {10}^{3.62\times {10}^{607}}$$$$\mathrm{maybe}\mathrm{try}\mathrm{to}\mathrm{find}\underset{n=1}{\sum ^{10}}{2}^{2^n}\mathrm{first}\mathrm{and}\mathrm{then}\mathrm{go}\mathrm{on}$$$$\mathrm{from}\mathrm{there}$$

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