Question Number 126952 by Study last updated on 25/Dec/20 $$\underset{{n}=\mathrm{1}} {\overset{\mathrm{10000}} {\sum}}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}=??? \\ $$ Commented by Study last updated on 25/Dec/20 $${help}\:{me} \\ $$ Answered…
Question Number 192481 by Tomal last updated on 19/May/23 Answered by aleks041103 last updated on 19/May/23 $$\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{1}+{x}}\left(\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)^{\mathrm{4}} +…\right) \\ $$$${if}\:\:\mid\frac{\mathrm{1}}{\mathrm{1}+{x}}\mid<\mathrm{1},\:{i}.{e}.\:\mid\mathrm{1}+{x}\mid>\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{1}+{x}}\:\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }}=\frac{\mathrm{1}+{x}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} −\mathrm{1}}…
Question Number 192477 by Shrinava last updated on 19/May/23 $$\mathrm{Find}:\:\:\:\:\:\frac{\frac{\mathrm{25}}{\mathrm{42}}\:−\:\frac{\mathrm{5}}{\mathrm{16}}\:+\:\frac{\mathrm{10}}{\mathrm{9}}\:−\:\frac{\mathrm{2}}{\mathrm{3}}}{\frac{\mathrm{3}}{\mathrm{8}}\:+\:\frac{\mathrm{4}}{\mathrm{5}}\:−\:\frac{\mathrm{5}}{\mathrm{7}}\:−\:\frac{\mathrm{4}}{\mathrm{3}}}\:=\:? \\ $$ Answered by Tomal last updated on 19/May/23 $$=\frac{\left(\frac{\mathrm{25}}{\mathrm{42}}−\frac{\mathrm{5}}{\mathrm{16}}\right)+\left(\frac{\mathrm{10}}{\mathrm{9}}−\frac{\mathrm{2}}{\mathrm{3}}\right)}{\left(\frac{\mathrm{315}+\mathrm{672}−\mathrm{600}−\mathrm{1120}}{\mathrm{840}}\right)} \\ $$$$=\frac{\left(\frac{\mathrm{200}−\mathrm{105}}{\mathrm{336}}\right)+\left(\frac{\mathrm{10}−\mathrm{6}}{\mathrm{9}}\right)}{−\frac{\mathrm{733}}{\mathrm{840}}} \\ $$$$=\frac{\frac{\mathrm{95}}{\mathrm{336}}+\frac{\mathrm{4}}{\mathrm{9}}}{−\frac{\mathrm{733}}{\mathrm{840}}} \\…
Question Number 126935 by Raxreedoroid last updated on 25/Dec/20 $${Prove}\:{or}\:{give}\:{a}\:{counter}\:{example}: \\ $$$$\left({a}+\mathrm{1}\right)^{{n}−\mathrm{1}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\underset{{k}−\mathrm{1}} {\overset{{n}−\mathrm{1}} {\:}}\right){a}^{{k}−\mathrm{1}} \\ $$$$\left(\underset{{r}} {\overset{{n}} {\:}}\right)=\frac{{n}!}{{r}!\left({n}−{r}\right)!} \\ $$ Terms of…
Question Number 126930 by bramlexs22 last updated on 25/Dec/20 $$\:\:\sqrt{\mathrm{1997}×\mathrm{1996}×\mathrm{1995}×\mathrm{1994}+\mathrm{1}}\:=? \\ $$ Commented by bramlexs22 last updated on 25/Dec/20 $$\Rightarrow{let}\:\mathrm{1994}={a}\:\rightarrow{a}\left({a}+\mathrm{1}\right)\left({a}+\mathrm{2}\right)\left({a}+\mathrm{3}\right)+\mathrm{1}= \\ $$$$\Leftrightarrow\left({a}^{\mathrm{2}} +{a}\right)\left({a}^{\mathrm{2}} +\mathrm{5}{a}+\mathrm{6}\right)+\mathrm{1}\:= \\…
Question Number 126931 by arash sharifi last updated on 25/Dec/20 $${y}=\left(\mathrm{1}−{x}\right)^{{cosx}} \\ $$ Answered by mahdipoor last updated on 25/Dec/20 $${salam}\:{arash}\:{jan},{soalat}\:{ro}\:{khamel}\:{benevis} \\ $$ Terms of…
Question Number 192463 by Spillover last updated on 18/May/23 Answered by AST last updated on 18/May/23 $$\mathrm{420}=\mathrm{2}^{\mathrm{2}} ×\mathrm{3}×\mathrm{5}×\mathrm{7} \\ $$$$\mathrm{264}=\mathrm{2}^{\mathrm{3}} ×\mathrm{3}×\mathrm{11} \\ $$$${Third}\:{number}=\mathrm{2}^{{a}} \mathrm{3}^{{b}} \mathrm{5}^{{c}}…
Question Number 192440 by cortano12 last updated on 18/May/23 $$\:\:\:\:\mathrm{Given}\:\begin{cases}{\mathrm{A}=\frac{\left(\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{pq}\right)^{\mathrm{2}} +\left(\mathrm{pr}\right)^{\mathrm{2}} +\left(\mathrm{qr}\right)^{\mathrm{2}} }}\\{\mathrm{B}=\frac{\mathrm{q}^{\mathrm{2}} −\mathrm{pr}}{\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} }\:}\end{cases}\:\:\:\:\:\: \\ $$$$\:\mathrm{If}\:\mathrm{p}+\mathrm{q}+\mathrm{r}=\mathrm{0}\:\mathrm{then}\:\mathrm{A}^{\mathrm{2}} −\mathrm{4B}=? \\…
Question Number 192437 by MATHEMATICSAM last updated on 18/May/23 $$\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}}\:=\:\frac{\mathrm{1}}{{a}\:+\:{b}\:+\:{c}}\:.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{5}} }\:+\:\frac{\mathrm{1}}{{b}^{\mathrm{5}} }\:+\:\frac{\mathrm{1}}{{c}^{\mathrm{5}} }\:=\:\frac{\mathrm{1}}{{a}^{\mathrm{5}} \:+\:{b}^{\mathrm{5}} \:+\:{c}^{\mathrm{5}} }\:=\:\frac{\mathrm{1}}{\left({a}\:+\:{b}\:+\:{c}\right)^{\mathrm{5}} } \\ $$ Answered by Frix last…
Question Number 192425 by mehdee42 last updated on 17/May/23 $${Question} \\ $$$${if}\:\:“{k}''\:{is}\:{odd}\:\:\&\:{A}=\mathrm{1}^{{k}} +\mathrm{2}^{{k}} +…+{n}^{{k}\:\:} \:\&\:\:{B}=\mathrm{1}+\mathrm{2}+…+{n} \\ $$$${prove}\:{that}\:\::\:\:{B}\:\mid\:{A}\: \\ $$ Answered by MM42 last updated on…