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Category: Arithmetic

Is-there-a-way-to-evaluate-the-following-S-2-2-2-2-2-

Question Number 2275 by Filup last updated on 13/Nov/15 $$\mathrm{Is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{way}\:\mathrm{to}\:\mathrm{evaluate}\:\mathrm{the}\:\mathrm{following}: \\ $$$$ \\ $$$${S}=\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+…}}}} \\ $$ Answered by Rasheed Soomro last updated on 13/Nov/15 $${S}=\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+…}}}}…

p-is-a-prime-number-such-that-1-p-p-2-7-find-all-k-such-that-p-k-42-

Question Number 67481 by Mohamed Amine Bouguezzoul last updated on 27/Aug/19 $${p}\:{is}\:{a}\:{prime}\:{number}\:{such}\:{that}\:\left(\mathrm{1}+{p}\right)^{{p}} \equiv\mathrm{2}\left[\mathrm{7}\right] \\ $$$${find}\:{all}\:{k}\:{such}\:{that}\:{p}\equiv{k}\left[\mathrm{42}\right] \\ $$ Commented by Rasheed.Sindhi last updated on 30/Aug/19 $$\boldsymbol{{Some}}\:{values}\:{of}\:{k}…

Solve-1-1-2-1-2-3-1-2-n-

Question Number 1831 by Filup last updated on 10/Oct/15 $$\mathrm{Solve}: \\ $$$$\mathrm{1}+\left(\mathrm{1}+\mathrm{2}\right)+\left(\mathrm{1}+\mathrm{2}+\mathrm{3}\right)+…+\left(\mathrm{1}+\mathrm{2}+…+{n}\right) \\ $$ Answered by Rasheed Soomro last updated on 10/Oct/15 $$\mathrm{1}+\left(\mathrm{1}+\mathrm{2}\right)+\left(\mathrm{1}+\mathrm{2}+\mathrm{3}\right)+…+\left(\mathrm{1}+\mathrm{2}+…+{n}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\right)\left(\mathrm{1}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\right)\left(\mathrm{2}+\mathrm{1}\right)+…+\frac{\mathrm{1}}{\mathrm{2}}{n}\left({n}+\mathrm{1}\right)…

1-1-1-2-2-3-1-4-1-5-2-6-1-7-1-8-2-9-1-10-1-11-2-12-

Question Number 66778 by Tony Lin last updated on 19/Aug/19 $$\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{2}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{2}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{11}}−\frac{\mathrm{2}}{\mathrm{12}}+\centerdot\centerdot\centerdot= \\ $$ Commented by Prithwish sen last updated on 19/Aug/19 $$\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+……\right)−\left[\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}\right)+\left(\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{2}}{\mathrm{6}}\right)+\left(\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{2}}{\mathrm{9}}\right)+\left(\frac{\mathrm{1}}{\mathrm{12}}+\frac{\mathrm{2}}{\mathrm{12}}\right)+…\right] \\ $$$$=\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty}…

2-2-

Question Number 1181 by 22 last updated on 11/Jul/15 $$\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}=? \\ $$ Answered by Rasheed Ahmad last updated on 24/Jul/15 $${In}\:{certain}\:{cases}\:\sqrt{{a}+{b}\sqrt{{c}}\:}\:{can}\:{be} \\ $$$${simplified}\:{into}\:{p}+{q}\sqrt{{c}}\:{form}\left({not}\right. \\ $$$$\left.{in}\:{all}\:{cases}\right).\:{The}\:{procedure}\:{is}\:{as}…