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Question Number 93091 by Maclaurin Stickker last updated on 10/May/20
Find the sum:  Σ_(n=1) ^(2020) 2^2^n
$${Find}\:{the}\:{sum}: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{2020}} {\sum}}\mathrm{2}^{\mathrm{2}^{{n}} } \\ $$
Commented by MJS last updated on 11/May/20
2^2^(2020)  ≈10^(3.62×10^(607) )   maybe try to find Σ_(n=1) ^(10) 2^2^n   first and then go on  from there
$$\mathrm{2}^{\mathrm{2}^{\mathrm{2020}} } \approx\mathrm{10}^{\mathrm{3}.\mathrm{62}×\mathrm{10}^{\mathrm{607}} } \\ $$$$\mathrm{maybe}\:\mathrm{try}\:\mathrm{to}\:\mathrm{find}\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\mathrm{2}^{\mathrm{2}^{{n}} } \:\mathrm{first}\:\mathrm{and}\:\mathrm{then}\:\mathrm{go}\:\mathrm{on} \\ $$$$\mathrm{from}\:\mathrm{there} \\ $$

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