Question Number 25122 by math solver last updated on 04/Dec/17 $$\:\mathrm{3}_{{C}_{\mathrm{1}} } \:+\:\mathrm{4}_{{C}_{\mathrm{2}} } \:+\:\mathrm{5}_{{C}_{\mathrm{3}} } \:+………..+\:\mathrm{49}_{{C}_{\mathrm{47}} } \:=\:? \\ $$$${where}\:{n}_{{C}_{{r}} } \:=\:\frac{{n}!}{{r}!×\left({n}−{r}\right)!}\:. \\ $$…
Question Number 90661 by Cynosure last updated on 25/Apr/20 $${show}\:{that}\:\left({n}^{\mathrm{4}} −{n}^{\mathrm{2}} \right)\:{is}\:{divisible}\:{by}\:\mathrm{12} \\ $$ Answered by MJS last updated on 25/Apr/20 $${n}^{\mathrm{4}} −{n}^{\mathrm{2}} ={n}^{\mathrm{2}} \left({n}^{\mathrm{2}}…
Question Number 25113 by Mr easy last updated on 04/Dec/17 Commented by prakash jain last updated on 05/Dec/17 $$\left(\mathrm{log}_{\mathrm{2}} {x}\right)^{\mathrm{3}} +\left(\mathrm{log}_{\mathrm{2}} {y}\right)^{\mathrm{3}} \\ $$$$=\left(\mathrm{log}_{\mathrm{2}} {x}+\mathrm{log}_{\mathrm{2}}…
Question Number 25114 by Mr easy last updated on 04/Dec/17 Commented by moxhix last updated on 04/Dec/17 $${x}=\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+{x}}}}>\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{0}}} \\ $$$$\therefore{x}>\sqrt{\mathrm{2}} \\ $$$${put}\:\phi:\:\:\phi=\sqrt{\mathrm{1}+\phi}\:\:\left(\phi>\mathrm{1}\right) \\ $$$$\phi^{\mathrm{2}} −\phi−\mathrm{1}=\mathrm{0}…
Question Number 25112 by Mr easy last updated on 04/Dec/17 Commented by prakash jain last updated on 04/Dec/17 $$\left({a}+{b}\right)^{\mathrm{2}{n}} =\underset{{i}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}\:^{\mathrm{2}{n}} {C}_{{i}} {a}^{{i}} {b}^{\mathrm{2}{n}−{i}}…
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Question Number 156176 by MathSh last updated on 08/Oct/21 Answered by ghimisi last updated on 09/Oct/21 $$\frac{{a}^{\mathrm{2}} }{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }+\frac{{c}^{\mathrm{2}} }{{z}^{\mathrm{2}} }=\mathrm{36} \\ $$$$\left.\mathrm{36}\centerdot\mathrm{4}=\left(\frac{{a}^{\mathrm{2}} }{{x}^{\mathrm{2}}…
Question Number 156172 by MathSh last updated on 08/Oct/21 Commented by Rasheed.Sindhi last updated on 09/Oct/21 $$\mathrm{Infinitely}\:\mathrm{many}\:\mathrm{sets}. \\ $$ Commented by MathSh last updated on…
Question Number 156170 by MathSh last updated on 08/Oct/21 Answered by mindispower last updated on 09/Oct/21 $${Hello}\:{sir}\:{withe}\:{resodue}\left[{theorm}\:{is}\:{diractly}\:\right. \\ $$$$\frac{{p}\left({x}\right)}{{q}\left({x}\right)}=\:{deg}\:{q}>{deg}\left({p}\right)+\mathrm{1}\:{in}\:{this}\:{case} \\ $$$$ \\ $$$$ \\ $$…
Question Number 156164 by MathSh last updated on 08/Oct/21 Answered by ghimisi last updated on 09/Oct/21 $$\mid{x}\mid=\frac{\mathrm{1}}{\mathrm{2}}\mid{x}+{y}+{x}−{y}\mid\leqslant\frac{\mathrm{1}}{\mathrm{2}}\mid{x}+{y}\mid+\frac{\mathrm{1}}{\mathrm{2}}\mid{y}−{x}\mid\:\:\left(\mathrm{1}\right) \\ $$$$\mid{y}\mid=\frac{\mathrm{1}}{\mathrm{2}}\mid{y}+{x}+{y}−{x}\mid\leqslant\frac{\mathrm{1}}{\mathrm{2}}\mid{x}+{y}\mid+\frac{\mathrm{1}}{\mathrm{2}}\mid{y}−{x}\mid\mid\:\:\left(\mathrm{2}\right) \\ $$$$\mid\mathrm{3}{x}+\mathrm{2}{y}\mid=\mid\mathrm{4}{x}+\mathrm{3}{y}−\left({x}+{y}\right)\mid\leqslant\mid\mathrm{4}{x}+\mathrm{3}{y}\mid+\mid{x}+{y}\mid\:\:\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)+\left(\mathrm{3}\right)\Rightarrow \\ $$$$\mid{x}\mid+\mid{y}\mid+\mid\mathrm{3}{x}+\mathrm{2}{y}\mid\leqslant\mid\mathrm{4}{x}+\mathrm{3}{y}\mid+\mathrm{2}\mid{x}+{y}\mid+\mid{y}−{x}\mid…