Question Number 153866 by amin96 last updated on 11/Sep/21
$$\mathrm{3}+\sqrt{\mathrm{3}+\sqrt{\mathrm{6}+\sqrt{\mathrm{9}+\sqrt{\mathrm{12}+\ldots+\sqrt{\mathrm{99}}}}}}=? \\ $$
Answered by peter frank last updated on 11/Sep/21
$$\mathrm{152122} \\ $$
Commented by amin96 last updated on 11/Sep/21
$$ \\ $$how did you do the solution sir?
Commented by mr W last updated on 11/Sep/21
$${i}\:{don}'{t}\:{think}\:{it}\:{can}\:{be}\:{solved}\:{in}\:{closed} \\ $$$${form}. \\ $$$${no}\:{matter}\:{how}\:{you}\:{got}\:{the}\:{result}\:\mathrm{152122}, \\ $$$${the}\:{result}\:{is}\:{wrong}. \\ $$$$ \\ $$$${S}=\mathrm{3}+\sqrt{\mathrm{3}+\sqrt{\mathrm{6}+\sqrt{\mathrm{9}+\sqrt{….+\sqrt{\mathrm{99}}}}}} \\ $$$$\:\:<\mathrm{3}+\sqrt{\mathrm{99}+\sqrt{\mathrm{99}+\sqrt{\mathrm{99}+\sqrt{….+\sqrt{\mathrm{99}}}}}} \\ $$$$\:\:\:<\mathrm{3}+\sqrt{\mathrm{99}+\sqrt{\mathrm{99}+\sqrt{\mathrm{99}+\sqrt{….+\sqrt{\mathrm{99}+\sqrt{…}}}}}} \\ $$$$\:\:\:=\mathrm{3}+{L} \\ $$$${with}\:{L}=\sqrt{\mathrm{99}+\sqrt{\mathrm{99}+\sqrt{\mathrm{99}+\sqrt{….+\sqrt{\mathrm{99}+\sqrt{…}}}}}} \\ $$$${L}=\sqrt{\mathrm{99}+{L}} \\ $$$${L}^{\mathrm{2}} −{L}−\mathrm{99}=\mathrm{0} \\ $$$${L}=\frac{\mathrm{1}+\sqrt{\mathrm{397}}}{\mathrm{2}}\approx\mathrm{10}.\mathrm{5} \\ $$$$\Rightarrow\mathrm{3}+\sqrt{\mathrm{3}+\sqrt{\mathrm{6}+\sqrt{\mathrm{9}+\sqrt{….+\sqrt{\mathrm{99}}}}}}\:<\:\mathrm{3}+{L}=\mathrm{13}.\mathrm{5} \\ $$$${so}\:{it}\:{can}\:{never}\:{be}\:\mathrm{152122}! \\ $$
Commented by MJS_new last updated on 11/Sep/21
$$\mathrm{can}\:\mathrm{we}\:\mathrm{find}\:\mathrm{a}\:\mathrm{closed}\:\mathrm{form}\:\mathrm{if}\:\mathrm{it}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{stop}? \\ $$$$\sqrt{\mathrm{1}×\mathrm{3}+\sqrt{\mathrm{2}×\mathrm{3}+\sqrt{\mathrm{3}×\mathrm{3}+…\sqrt{{n}×\mathrm{3}}}}}\:\mathrm{with}\:{n}\rightarrow\infty \\ $$$$\mathrm{or}\:\mathrm{at}\:\mathrm{least}\:\mathrm{for} \\ $$$$\sqrt{\mathrm{1}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}+\sqrt{\mathrm{4}+…\sqrt{{n}}}}}}\:\mathrm{with}\:{n}\rightarrow\infty \\ $$