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Question Number 194144 by mr W last updated on 28/Jun/23
fill with different natural numbers: (1/(19))=(1/(( )))+(1/(( )))+(1/(( )))+(1/(( )))
$${fill}\:{with}\:{different}\:{natural}\:{numbers}: \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{19}}=\frac{\mathrm{1}}{\left(\:\:\right)}+\frac{\mathrm{1}}{\left(\:\:\right)}+\frac{\mathrm{1}}{\left(\:\:\right)}+\frac{\mathrm{1}}{\left(\:\:\right)} \\ $$
Answered by York12 last updated on 28/Jun/23
egyptian fractions (1/k)=(1/(k+1))+(1/(k(k+1)))
$${egyptian}\:{fractions} \\ $$$$\frac{\mathrm{1}}{{k}}=\frac{\mathrm{1}}{{k}+\mathrm{1}}+\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)} \\ $$
Answered by BaliramKumar last updated on 28/Jun/23
(1/(19)) = (1/((20))) + (1/((381))) + (1/((144781))) + (1/((20961393180)))
$$\frac{\mathrm{1}}{\mathrm{19}}\:=\:\frac{\mathrm{1}}{\left(\mathrm{20}\right)}\:+\:\frac{\mathrm{1}}{\left(\mathrm{381}\right)}\:+\:\frac{\mathrm{1}}{\left(\mathrm{144781}\right)}\:+\:\frac{\mathrm{1}}{\left(\mathrm{20961393180}\right)}\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by Frix last updated on 28/Jun/23
(1/(21))+(1/(381))+(1/(420))+(1/(144780))
$$\frac{\mathrm{1}}{\mathrm{21}}+\frac{\mathrm{1}}{\mathrm{381}}+\frac{\mathrm{1}}{\mathrm{420}}+\frac{\mathrm{1}}{\mathrm{144780}} \\ $$
Commented by BaliramKumar last updated on 28/Jun/23
Also (1/(19)) = (1/(19×20)) + (1/(20×21)) + (1/(21×22)) + (1/(22)) (1/(19)) = (1/((22))) + (1/((380))) + (1/((420))) + (1/((462))) determinant ((((1/n) = (1/(n(n+1))) + (1/((n+1)(n+2))) + (1/((n+2)(n+3))) + (1/((n+3))) ......)))
$$\mathrm{Also} \\ $$$$\frac{\mathrm{1}}{\mathrm{19}}\:=\:\frac{\mathrm{1}}{\mathrm{19}×\mathrm{20}}\:+\:\frac{\mathrm{1}}{\mathrm{20}×\mathrm{21}}\:+\:\frac{\mathrm{1}}{\mathrm{21}×\mathrm{22}}\:+\:\frac{\mathrm{1}}{\mathrm{22}}\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{19}}\:=\:\frac{\mathrm{1}}{\left(\mathrm{22}\right)}\:+\:\frac{\mathrm{1}}{\left(\mathrm{380}\right)}\:+\:\frac{\mathrm{1}}{\left(\mathrm{420}\right)}\:+\:\frac{\mathrm{1}}{\left(\mathrm{462}\right)}\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\begin{array}{|c|}{\frac{\mathrm{1}}{\mathrm{n}}\:=\:\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}\:+\:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)}\:+\:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{3}\right)}\:+\:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{3}\right)}\:……}\\\hline\end{array}\:\:\:\:\:\: \\ $$$$ \\ $$
Answered by witcher3 last updated on 28/Jun/23
(1/(19))=(1/(19)).1 1=((10)/(10))=((5+2+2+1)/(10))=(1/2)+(1/5)+(1/5)+(1/(10))
$$\frac{\mathrm{1}}{\mathrm{19}}=\frac{\mathrm{1}}{\mathrm{19}}.\mathrm{1} \\ $$$$\mathrm{1}=\frac{\mathrm{10}}{\mathrm{10}}=\frac{\mathrm{5}+\mathrm{2}+\mathrm{2}+\mathrm{1}}{\mathrm{10}}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{10}} \\ $$$$ \\ $$
Commented by BaliramKumar last updated on 28/Jun/23
Nice approach but(1/5) & (1/5) same number (1/(19)) = (1/(19))×((20)/(20)) = (1/(19))(((10+5+4+1)/(20))) (1/(19)) = (1/(19))((1/2) + (1/4) + (1/5) + (1/(20))) (1/(19)) = (1/(2×19)) + (1/(4×19)) + (1/(5×19)) + (1/(20×19)) (1/(19)) = (1/(38)) + (1/(76)) + (1/(95)) + (1/(380))
$$\mathrm{Nice}\:\mathrm{approach}\:\mathrm{but}\frac{\mathrm{1}}{\mathrm{5}}\:\&\:\frac{\mathrm{1}}{\mathrm{5}}\:\mathrm{same}\:\mathrm{number} \\ $$$$\frac{\mathrm{1}}{\mathrm{19}}\:=\:\frac{\mathrm{1}}{\mathrm{19}}×\frac{\mathrm{20}}{\mathrm{20}}\:=\:\frac{\mathrm{1}}{\mathrm{19}}\left(\frac{\mathrm{10}+\mathrm{5}+\mathrm{4}+\mathrm{1}}{\mathrm{20}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{19}}\:=\:\frac{\mathrm{1}}{\mathrm{19}}\left(\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{4}}\:+\:\frac{\mathrm{1}}{\mathrm{5}}\:+\:\frac{\mathrm{1}}{\mathrm{20}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{19}}\:=\:\frac{\mathrm{1}}{\mathrm{2}×\mathrm{19}}\:+\:\frac{\mathrm{1}}{\mathrm{4}×\mathrm{19}}\:+\:\frac{\mathrm{1}}{\mathrm{5}×\mathrm{19}}\:+\:\frac{\mathrm{1}}{\mathrm{20}×\mathrm{19}} \\ $$$$\frac{\mathrm{1}}{\mathrm{19}}\:=\:\frac{\mathrm{1}}{\mathrm{38}}\:+\:\frac{\mathrm{1}}{\mathrm{76}}\:+\:\frac{\mathrm{1}}{\mathrm{95}}\:+\:\frac{\mathrm{1}}{\mathrm{380}} \\ $$$$ \\ $$
Commented by York12 last updated on 28/Jun/23
there are a lot actually as I have mentioned above it is an extension to the egyptian fractions which are still a mystrey till today
$${there}\:{are}\:{a}\:{lot}\:{actually}\:{as}\:{I}\:{have}\:{mentioned} \\ $$$${above}\:{it}\:{is}\:{an}\:{extension}\:{to}\: \\ $$$${the}\:{egyptian}\:{fractions}\:{which}\:{are}\:{still} \\ $$$${a}\:{mystrey}\:{till}\:{today} \\ $$
Commented by MM42 last updated on 28/Jun/23
if d_i ∣m ; 1≤i≤k & Σ d_i =n ⇒((Σd_i )/m)=(1/m_1 )+(1/m_2 )+...+(1/m_k ) ; m_i =(m/d_i ) for example: (1/(24))=((12+8+3+1)/(24))=(1/(12))+(1/3)+(1/8)+(1/(24)) or (1/(24))=((8+4+6+3+2+1)/(24))=(1/3)+(1/6)+(1/4)+(1/8)+(1/(24))
$${if}\:\:\:\:{d}_{{i}} \:\mid{m}\:\:\:;\:\:\mathrm{1}\leqslant{i}\leqslant{k}\:\:\:\&\:\:\Sigma\:{d}_{{i}} ={n} \\ $$$$\Rightarrow\frac{\Sigma{d}_{{i}} }{{m}}=\frac{\mathrm{1}}{{m}_{\mathrm{1}} }+\frac{\mathrm{1}}{{m}_{\mathrm{2}} }+…+\frac{\mathrm{1}}{{m}_{{k}} }\:\:\:;\:\:{m}_{{i}} =\frac{{m}}{{d}_{{i}} } \\ $$$${for}\:\:{example}: \\ $$$$\frac{\mathrm{1}}{\mathrm{24}}=\frac{\mathrm{12}+\mathrm{8}+\mathrm{3}+\mathrm{1}}{\mathrm{24}}=\frac{\mathrm{1}}{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{24}} \\ $$$${or}\:\:\frac{\mathrm{1}}{\mathrm{24}}=\frac{\mathrm{8}+\mathrm{4}+\mathrm{6}+\mathrm{3}+\mathrm{2}+\mathrm{1}}{\mathrm{24}}=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{24}} \\ $$$$ \\ $$
Answered by Spillover last updated on 01/Jul/23
By using Egptian fraction.we can express (1/(19)) as sum of unit fraction (1/(19))=(1/((3)))+(1/((37)))+(1/((228)))+(1/((342)))
$${By}\:{using}\:{Egptian}\:{fraction}.{we}\:{can}\:{express} \\ $$$$\frac{\mathrm{1}}{\mathrm{19}}\:{as}\:{sum}\:{of}\:{unit}\:{fraction} \\ $$$$\frac{\mathrm{1}}{\mathrm{19}}=\frac{\mathrm{1}}{\left(\mathrm{3}\right)}+\frac{\mathrm{1}}{\left(\mathrm{37}\right)}+\frac{\mathrm{1}}{\left(\mathrm{228}\right)}+\frac{\mathrm{1}}{\left(\mathrm{342}\right)} \\ $$
Answered by Spillover last updated on 01/Jul/23

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