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Category: Arithmetic

challenge-question-Let-R-m-n-R-m-n-1-R-m-1-n-1-if-R-1-1-19-R-2-1-9-R-3-1-2-and-n-s-t-R-4-n-0-a-Find-R-1-33-b-Find-R-1-n-

Question Number 136005 by Raxreedoroid last updated on 17/Mar/21 $$\mathrm{challenge}\:\mathrm{question} \\ $$$$ \\ $$$$\mathrm{Let}\:{R}\left({m},{n}\right)={R}\left({m},{n}−\mathrm{1}\right)+{R}\left({m}+\mathrm{1},{n}−\mathrm{1}\right) \\ $$$$\mathrm{if}\:{R}\left(\mathrm{1},\mathrm{1}\right)=\mathrm{19},{R}\left(\mathrm{2},\mathrm{1}\right)=\mathrm{9},{R}\left(\mathrm{3},\mathrm{1}\right)=−\mathrm{2} \\ $$$$\mathrm{and}\:\forall{n}\:\mathrm{s}.\mathrm{t}.\:{R}\left(\mathrm{4},{n}\right)=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$\left.{a}\right)\:\mathrm{Find}\:{R}\left(\mathrm{1},\mathrm{33}\right) \\…

Question-70417

Question Number 70417 by aliesam last updated on 04/Oct/19 Answered by MJS last updated on 04/Oct/19 $$\mathrm{just}\:\mathrm{use}\:\mathrm{formulas} \\ $$$$\mathrm{sphere}:\:{V}=\frac{\mathrm{4}\pi}{\mathrm{3}}{r}^{\mathrm{3}} ;\:{S}=\mathrm{4}\pi{r}^{\mathrm{2}} \\ $$$$\mathrm{cylinder}:\:{V}=\pi{r}^{\mathrm{2}} {h} \\ $$$$\:\:\:\:\:{S}=\mathrm{2}×\mathrm{circle}+\mathrm{lateral}=\mathrm{2}\pi{r}^{\mathrm{2}}…

Solve-x-4-x-3-2ax-2-ax-a-2-0-a-R-

Question Number 70399 by Henri Boucatchou last updated on 04/Oct/19 $$\boldsymbol{{Solve}}\:\:\boldsymbol{{x}}^{\mathrm{4}} \:+\:\boldsymbol{{x}}^{\mathrm{3}} \:−\mathrm{2}\boldsymbol{{ax}}^{\mathrm{2}} \:−\boldsymbol{{ax}}\:+\:\boldsymbol{{a}}^{\mathrm{2}} =\:\mathrm{0},\:\:\boldsymbol{{a}}\:\in\:\mathbb{R} \\ $$ Commented by Prithwish sen last updated on 04/Oct/19…

1-If-A-and-B-are-sets-define-their-scheffer-product-A-B-by-A-B-A-B-Prove-by-definitions-that-A-B-A-B-A-B-2-State-the-strong-principle-of-mathematical-induction-Suppose-that-a-1-1-a-2-3-a

Question Number 135889 by Ar Brandon last updated on 16/Mar/21 $$\mathrm{1}\backslash\:\mathrm{If}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{sets}\:\mathrm{define}\:\mathrm{their}\:\mathrm{scheffer}\:\mathrm{product}\:\mathrm{A}\ast\mathrm{B}\:\mathrm{by}\:\mathrm{A}\ast\mathrm{B}=\mathrm{A}\ast\cap\mathrm{B}\ast \\ $$$$\mathrm{Prove}\:\mathrm{by}\:\mathrm{definitions}\:\mathrm{that}\:\left(\mathrm{A}\ast\mathrm{B}\right)\ast\left(\mathrm{A}\ast\mathrm{B}\right)=\mathrm{A}\cup\mathrm{B} \\ $$$$ \\ $$$$\mathrm{2}\backslash\:\mathrm{State}\:\mathrm{the}\:\mathrm{strong}\:\mathrm{principle}\:\mathrm{of}\:\mathrm{mathematical}\:\mathrm{induction}. \\ $$$$\mathrm{Suppose}\:\mathrm{that}\:\mathrm{a}_{\mathrm{1}} =\mathrm{1}\:,\:\mathrm{a}_{\mathrm{2}} =\mathrm{3} \\ $$$$\mathrm{a}_{\mathrm{k}} =\mathrm{a}_{\mathrm{k}−\mathrm{2}} +\mathrm{2a}_{\mathrm{k}−\mathrm{1}}…

If-a-1-a-2-a-n-be-an-arithmetic-progression-then-show-that-1-a-1-a-n-1-a-2-a-n-1-1-a-3-a-n-2-1-a-n-a-1-2-a-1-a-n-1-a-1-1-a-2-

Question Number 4704 by lakshaysethi039 last updated on 25/Feb/16 $${If}\:{a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,………{a}_{{n}} {be}\:{an}\:{arithmetic}\:{progression}, \\ $$$${then}\:{show}\:{that} \\ $$$$\frac{\mathrm{1}}{{a}_{\mathrm{1}} {a}_{{n}} }\:+\:\frac{\mathrm{1}}{{a}_{\mathrm{2}} {a}_{{n}−\mathrm{1}} }\:+\:\frac{\mathrm{1}}{{a}_{\mathrm{3}} {a}_{{n}−\mathrm{2}} }\:+………….+\frac{\mathrm{1}}{{a}_{{n}} {a}_{\mathrm{1}} }\:…

Is-the-section-marked-correct-i-1-x-i-x-n-1-x-i-1-n-i-1-2-x-1-1-2-1-2-3-1-2-x-1-x-2-x-1-3-x-2-x-1-2-x-1-m-1-x-n-1-m-i-1-

Question Number 4639 by FilupSmith last updated on 17/Feb/16 $$\mathrm{Is}\:\mathrm{the}\:\mathrm{section}\:\mathrm{marked}\:\left(\ast\right)\:{correct}? \\ $$$$ \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{x}} {\prod}}{i}={x}! \\ $$$$ \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{x}} {\prod}}\left(\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}{i}\right)=\mathrm{1}!×\mathrm{2}!×…×{x}! \\…