Question Number 111610 by A8;15: last updated on 04/Sep/20 | ||
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Commented by A8;15: last updated on 04/Sep/20 | ||
please help | ||
Commented by A8;15: last updated on 04/Sep/20 | ||
please | ||
Answered by Her_Majesty last updated on 04/Sep/20 | ||
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$$\frac{{b}^{{n}+\mathrm{1}} +{b}^{{n}} +\mathrm{1}}{{b}^{{n}} +\mathrm{1}}={b}+\mathrm{1}−\frac{{b}}{{b}^{{n}} +\mathrm{1}} \\ $$$${with}\:{b}=\mathrm{7}\:{and}\:{n}=\mathrm{2004}\:{we}\:{get} \\ $$$$\mathrm{8}−\frac{\mathrm{7}}{\mathrm{7}^{\mathrm{2004}} +\mathrm{1}}\approx\mathrm{8}−\mathrm{10}^{−\mathrm{1693}} \approx\mathrm{8} \\ $$ | ||