Question Number 64955 by Tony Lin last updated on 23/Jul/19 $${prove}\:{that}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\centerdot\frac{\left({k}+\mathrm{1}\right)!}{\mathrm{2}^{{k}+\mathrm{1}} }=\frac{\left({n}+\mathrm{2}\right)!}{\mathrm{2}^{{n}+\mathrm{1}} }−\mathrm{1} \\ $$ Commented by ~ À ® @ 237 ~…
Question Number 64916 by Masumsiddiqui399@gmail.com last updated on 23/Jul/19 Commented by Tawa1 last updated on 23/Jul/19 $$\left(\mathrm{2}\:+\:\sqrt{\mathrm{3}}\right)^{\mathrm{x}} \:+\:\left(\frac{\mathrm{1}}{\mathrm{2}\:+\:\sqrt{\mathrm{3}}}\right)^{\mathrm{x}} \:=\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\mathrm{since}\:\:\:\:\mathrm{2}\:−\:\sqrt{\mathrm{3}}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{2}\:+\:\sqrt{\mathrm{3}}}\:\:\mathrm{on}\:\mathrm{rationalise}\right] \\ $$$$\left(\mathrm{2}\:+\:\sqrt{\mathrm{3}}\right)^{\mathrm{x}} \:+\:\frac{\mathrm{1}}{\left(\mathrm{2}\:+\:\sqrt{\mathrm{3}}\right)^{\mathrm{x}} }\:=\:\:\mathrm{4} \\ $$$$\mathrm{Let}\:\:\:\:\:\left(\mathrm{2}\:+\:\sqrt{\mathrm{3}}\right)^{\mathrm{x}}…
Question Number 64564 by Tawa1 last updated on 19/Jul/19 Commented by Tony Lin last updated on 19/Jul/19 $${let}\:{cosx}={t} \\ $$$$\frac{{t}}{\mathrm{1}+\frac{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{{t}}}−\frac{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{\mathrm{1}+{t}} \\ $$$$=\frac{{t}^{\mathrm{2}} }{\mathrm{1}+\sqrt{\mathrm{1}−{t}^{\mathrm{2}}…
Question Number 64561 by aliesam last updated on 19/Jul/19 Commented by mathmax by abdo last updated on 20/Jul/19 $${let}\:{P}_{{n}} =\prod_{{k}=\mathrm{0}} ^{{n}} \left(\mathrm{1}+\sigma^{\mathrm{2}^{{k}} } \right)\:\Rightarrow{P}_{{n}} =\left(\mathrm{1}+\sigma\right)\left(\mathrm{1}+\sigma^{\mathrm{2}}…
Question Number 130085 by liberty last updated on 22/Jan/21 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{if}\:\begin{cases}{\mathrm{a}=\mathrm{r}^{\mathrm{2}} −\mathrm{2rs}−\mathrm{s}^{\mathrm{2}} }\\{\mathrm{b}=\mathrm{r}^{\mathrm{2}} +\mathrm{s}^{\mathrm{2}} }\\{\mathrm{c}=\mathrm{r}^{\mathrm{2}} +\mathrm{2rs}−\mathrm{s}^{\mathrm{2}} }\end{cases} \\ $$$$\:\mathrm{for}\:\mathrm{some}\:\mathrm{integers}\:\mathrm{r},\mathrm{s}\:\mathrm{then}\:\mathrm{a}^{\mathrm{2}} ,\mathrm{b}^{\mathrm{2}} ,\mathrm{c}^{\mathrm{2}} \\ $$$$\mathrm{are}\:\mathrm{three}\:\mathrm{square}\:\mathrm{in}\:\mathrm{AP}. \\ $$ Answered…
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Question Number 64153 by meme last updated on 14/Jul/19 $${solve}\:{to}\:{z}^{\mathrm{2}} \:\:\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{24} \\ $$ Answered by mr W last updated on 15/Jul/19 $$\left({x}−{y}\right)\left({x}+{y}\right)=\mathrm{24}={a}×{b}=\mathrm{1}×\mathrm{24}=\mathrm{2}×\mathrm{12}=\mathrm{3}×\mathrm{8}=\mathrm{4}×\mathrm{6} \\…
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