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Question Number 113059 by bemath last updated on 11/Sep/20

Answered by john santu last updated on 11/Sep/20

(1/(2.3.4)) + (1/(3.4.5))+(1/(4.5.6))+(1/(5.6.7)) = Σ_(k=1) ^4 (1/((k+1).(k+2).(k+3))) (1/((k+1)(k+2)(k+3)))=(p/(k+1))+(q/(k+2))+(r/(k+3)) 1=p(k+2)(k+3)+q(k+1)(k+3)+r(k+1)(k+2) k=−1⇒1=p(1)(2); p=(1/2) k=−2⇒1=q(−1)(1); q=−1 k=−3⇒1=r(−2)(−1);r=(1/2) so we have Σ_(k=1) ^4 (1/(2(k+1)))+(1/(2(k+3)))−(1/((k+2))) = (1/4)+(1/8)−(1/3)+(1/6)+(1/(10))−(1/4)+(1/8)+(1/(12))−(1/5) +(1/(10))+(1/(14))−(1/6) =(1/4)+(1/(12))+(1/(14))−(1/3)=((7+6)/(84))−(1/(12))=((13)/(84))−(7/(84)) = (6/(84)) = (1/(14))

$$\:\frac{\mathrm{1}}{\mathrm{2}.\mathrm{3}.\mathrm{4}}\:+\:\frac{\mathrm{1}}{\mathrm{3}.\mathrm{4}.\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{4}.\mathrm{5}.\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{5}.\mathrm{6}.\mathrm{7}}\:= \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right).\left({k}+\mathrm{2}\right).\left({k}+\mathrm{3}\right)}\: \\ $$$$\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)\left({k}+\mathrm{3}\right)}=\frac{{p}}{{k}+\mathrm{1}}+\frac{{q}}{{k}+\mathrm{2}}+\frac{{r}}{{k}+\mathrm{3}} \\ $$$$\mathrm{1}={p}\left({k}+\mathrm{2}\right)\left({k}+\mathrm{3}\right)+{q}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{3}\right)+{r}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right) \\ $$$${k}=−\mathrm{1}\Rightarrow\mathrm{1}={p}\left(\mathrm{1}\right)\left(\mathrm{2}\right);\:{p}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${k}=−\mathrm{2}\Rightarrow\mathrm{1}={q}\left(−\mathrm{1}\right)\left(\mathrm{1}\right);\:{q}=−\mathrm{1} \\ $$$${k}=−\mathrm{3}\Rightarrow\mathrm{1}={r}\left(−\mathrm{2}\right)\left(−\mathrm{1}\right);{r}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${so}\:{we}\:{have}\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2}\left({k}+\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}\left({k}+\mathrm{3}\right)}−\frac{\mathrm{1}}{\left({k}+\mathrm{2}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{10}}−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$+\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{14}}−\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{14}}−\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{7}+\mathrm{6}}{\mathrm{84}}−\frac{\mathrm{1}}{\mathrm{12}}=\frac{\mathrm{13}}{\mathrm{84}}−\frac{\mathrm{7}}{\mathrm{84}} \\ $$$$=\:\frac{\mathrm{6}}{\mathrm{84}}\:=\:\frac{\mathrm{1}}{\mathrm{14}} \\ $$

Answered by floor(10²Eta[1]) last updated on 11/Sep/20

(1/(2.3.4))+(1/(3.4.5))+(1/(4.5.6))+(1/(5.6.7)) =(1/(4!))+((2!)/(5!))+((3!)/(6!))+((4!)/(7!)) =(1/(4!))(1+((2!)/5)+((3!)/(6.5))+((4!)/(7.6.5))) =(1/(4!))(1+(2/5)+(1/5)+(4/(7.5))) =(1/(4!))(((60)/(5.7)))=(1/4)×((10)/(5.7))=(1/2)×(1/7)=(1/(14))

$$\frac{\mathrm{1}}{\mathrm{2}.\mathrm{3}.\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{3}.\mathrm{4}.\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{4}.\mathrm{5}.\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{5}.\mathrm{6}.\mathrm{7}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}!}+\frac{\mathrm{2}!}{\mathrm{5}!}+\frac{\mathrm{3}!}{\mathrm{6}!}+\frac{\mathrm{4}!}{\mathrm{7}!} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}!}\left(\mathrm{1}+\frac{\mathrm{2}!}{\mathrm{5}}+\frac{\mathrm{3}!}{\mathrm{6}.\mathrm{5}}+\frac{\mathrm{4}!}{\mathrm{7}.\mathrm{6}.\mathrm{5}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}!}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{4}}{\mathrm{7}.\mathrm{5}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}!}\left(\frac{\mathrm{60}}{\mathrm{5}.\mathrm{7}}\right)=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{10}}{\mathrm{5}.\mathrm{7}}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{7}}=\frac{\mathrm{1}}{\mathrm{14}} \\ $$

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