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Question Number 151426 by peter frank last updated on 21/Aug/21

Express (1/(5×9)) in partial fraction

$$\mathrm{Express}\:\frac{\mathrm{1}}{\mathrm{5}×\mathrm{9}}\:\mathrm{in}\:\mathrm{partial}\:\mathrm{fraction} \\ $$

Answered by liberty last updated on 21/Aug/21

(1/(n(n+4)))=(a/n)+(b/(n+4)) a= [(1/(n+4)) ]_(n=0) =(1/4) b= [(1/n) ]_(n=−4) =−(1/4) (1/(n(n+4)))=(1/(4n))−(1/(4(n+4))) n=5⇒(1/(5×9)) =(1/(20))−(1/(36))

$$\:\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{4}\right)}=\frac{\mathrm{a}}{\mathrm{n}}+\frac{\mathrm{b}}{\mathrm{n}+\mathrm{4}}\: \\ $$$$\mathrm{a}=\:\left[\frac{\mathrm{1}}{\mathrm{n}+\mathrm{4}}\:\right]_{\mathrm{n}=\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{b}=\:\left[\frac{\mathrm{1}}{\mathrm{n}}\:\right]_{\mathrm{n}=−\mathrm{4}} =−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{4}\right)}=\frac{\mathrm{1}}{\mathrm{4n}}−\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{n}+\mathrm{4}\right)} \\ $$$$\mathrm{n}=\mathrm{5}\Rightarrow\frac{\mathrm{1}}{\mathrm{5}×\mathrm{9}}\:=\frac{\mathrm{1}}{\mathrm{20}}−\frac{\mathrm{1}}{\mathrm{36}}\: \\ $$

Answered by MJS_new last updated on 21/Aug/21

it′s not unique (1/((a−b)(a+b)))=(1/(2(a−b)b))−(1/(2(a+b)b))= [a=7∧b=2] =(1/(20))−(1/(36)) but (1/((a−b)(a+b)))=(1/(2a(a+b)))+(1/(2a(a−b)))= [a=7∧b=2] =(1/(126))+(1/(70))

$$\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{unique} \\ $$$$\frac{\mathrm{1}}{\left({a}−{b}\right)\left({a}+{b}\right)}=\frac{\mathrm{1}}{\mathrm{2}\left({a}−{b}\right){b}}−\frac{\mathrm{1}}{\mathrm{2}\left({a}+{b}\right){b}}= \\ $$$$\:\:\:\:\:\left[{a}=\mathrm{7}\wedge{b}=\mathrm{2}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{20}}−\frac{\mathrm{1}}{\mathrm{36}} \\ $$$$\mathrm{but} \\ $$$$\frac{\mathrm{1}}{\left({a}−{b}\right)\left({a}+{b}\right)}=\frac{\mathrm{1}}{\mathrm{2}{a}\left({a}+{b}\right)}+\frac{\mathrm{1}}{\mathrm{2}{a}\left({a}−{b}\right)}= \\ $$$$\:\:\:\:\:\left[{a}=\mathrm{7}\wedge{b}=\mathrm{2}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{126}}+\frac{\mathrm{1}}{\mathrm{70}} \\ $$

Answered by MJS_new last updated on 21/Aug/21

(1/(5×9))=(1/a)+(1/b) with a≥b ((a),(b) ) ∈{ (((18)),((−30)) ) , (((20)),((−36)) ) , (((30)),((−90)) ) , (((36)),((−180)) ) , (((40)),((−360)) ) , (((42)),((−630)) ) , (((44)),((−1980)) ) , (((90)),((90)) ) , (((120)),((72)) ) , (((126)),((70)) ) , (((180)),((60)) ) , (((270)),((54)) ) , (((450)),((50)) ) , (((720)),((48)) ) , (((2070)),((46)) ) }

$$\frac{\mathrm{1}}{\mathrm{5}×\mathrm{9}}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\:\mathrm{with}\:{a}\geqslant{b} \\ $$$$\begin{pmatrix}{{a}}\\{{b}}\end{pmatrix}\:\in\left\{\begin{pmatrix}{\mathrm{18}}\\{−\mathrm{30}}\end{pmatrix}\:,\:\begin{pmatrix}{\mathrm{20}}\\{−\mathrm{36}}\end{pmatrix}\:,\:\begin{pmatrix}{\mathrm{30}}\\{−\mathrm{90}}\end{pmatrix}\:,\:\begin{pmatrix}{\mathrm{36}}\\{−\mathrm{180}}\end{pmatrix}\:,\:\begin{pmatrix}{\mathrm{40}}\\{−\mathrm{360}}\end{pmatrix}\:,\:\begin{pmatrix}{\mathrm{42}}\\{−\mathrm{630}}\end{pmatrix}\:,\:\begin{pmatrix}{\mathrm{44}}\\{−\mathrm{1980}}\end{pmatrix}\:,\:\begin{pmatrix}{\mathrm{90}}\\{\mathrm{90}}\end{pmatrix}\:,\:\begin{pmatrix}{\mathrm{120}}\\{\mathrm{72}}\end{pmatrix}\:,\:\begin{pmatrix}{\mathrm{126}}\\{\mathrm{70}}\end{pmatrix}\:,\:\begin{pmatrix}{\mathrm{180}}\\{\mathrm{60}}\end{pmatrix}\:,\:\begin{pmatrix}{\mathrm{270}}\\{\mathrm{54}}\end{pmatrix}\:,\:\begin{pmatrix}{\mathrm{450}}\\{\mathrm{50}}\end{pmatrix}\:,\:\begin{pmatrix}{\mathrm{720}}\\{\mathrm{48}}\end{pmatrix}\:,\:\begin{pmatrix}{\mathrm{2070}}\\{\mathrm{46}}\end{pmatrix}\:\right\} \\ $$

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