Menu Close

Find-2-35-2-63-2-99-2-143-

Tinku Tara 0 Comments
Pin




Question Number 201430 by hardmath last updated on 06/Dec/23
Find: (2/(35)) + (2/(63)) + (2/(99)) + (2/(143)) = ?
$$\mathrm{Find}: \\ $$$$\frac{\mathrm{2}}{\mathrm{35}}\:+\:\frac{\mathrm{2}}{\mathrm{63}}\:+\:\frac{\mathrm{2}}{\mathrm{99}}\:+\:\frac{\mathrm{2}}{\mathrm{143}}\:=\:? \\ $$
Answered by Rasheed.Sindhi last updated on 06/Dec/23
2((1/(6^2 −1))+(1/(8^2 −1))+(1/(10^2 −1))+(1/(12^2 −1))) 2((1/((2(1)+4)^2 −1))+(1/((2(2)+4)^2 −1))+(1/((2(3)+4)^2 −1))+(1/((2(4)+4)^2 −1))) General term=(1/((2n+4)^2 −1)) =(1/((2n+4−1)(2n+4+1)))=(1/((2n+3)(2n+5))) Partial fraction: (a/(2n+3))+(b/(2n+5))=(1/((2n+3)(2n+5))) a(2n+5)+b(2n+3)=1 2an+2nb+5a+3b=1 (2a+2b)n+(5a+3b)=1 2a+2b=0 ∧ 5a+3b=1 b=−a ∧ 5a−3a=1⇒a=(1/2)⇒b=−(1/2) (1/((2n+3)(2n+5)))=((1/2)/(2n+3))+((−1/2)/(2n+5))=(1/2)((1/(2n+3))−(1/(2n+5))) =2Σ_(n=1) ^4 {(1/2)((1/(2n+3))−(1/(2n+5)))} =Σ_(n=1) ^4 ((1/(2n+3))−(1/(2n+5))) =Σ_(n=1) ^4 ((1/(2n+3)))−Σ_(n=1) ^4 ((1/(2n+5))) =((1/5)−(1/7))+((1/7)−(1/9))+((1/9)−(1/(11)))+((1/(11))−(1/(13))) =(1/5)−(1/(13))=(8/(65))
$$\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{8}^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{10}^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{12}^{\mathrm{2}} −\mathrm{1}}\right) \\ $$$$\mathrm{2}\left(\frac{\mathrm{1}}{\left(\mathrm{2}\left(\mathrm{1}\right)+\mathrm{4}\right)^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{\left(\mathrm{2}\left(\mathrm{2}\right)+\mathrm{4}\right)^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{\left(\mathrm{2}\left(\mathrm{3}\right)+\mathrm{4}\right)^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{\left(\mathrm{2}\left(\mathrm{4}\right)+\mathrm{4}\right)^{\mathrm{2}} −\mathrm{1}}\right) \\ $$$$\mathcal{G}{eneral}\:{term}=\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{4}\right)^{\mathrm{2}} −\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{4}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{4}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{3}\right)\left(\mathrm{2}{n}+\mathrm{5}\right)} \\ $$$$\mathcal{P}{artial}\:{fraction}: \\ $$$$\frac{{a}}{\mathrm{2}{n}+\mathrm{3}}+\frac{{b}}{\mathrm{2}{n}+\mathrm{5}}=\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{3}\right)\left(\mathrm{2}{n}+\mathrm{5}\right)} \\ $$$${a}\left(\mathrm{2}{n}+\mathrm{5}\right)+{b}\left(\mathrm{2}{n}+\mathrm{3}\right)=\mathrm{1} \\ $$$$\mathrm{2}{an}+\mathrm{2}{nb}+\mathrm{5}{a}+\mathrm{3}{b}=\mathrm{1} \\ $$$$\left(\mathrm{2}{a}+\mathrm{2}{b}\right){n}+\left(\mathrm{5}{a}+\mathrm{3}{b}\right)=\mathrm{1} \\ $$$$\mathrm{2}{a}+\mathrm{2}{b}=\mathrm{0}\:\wedge\:\mathrm{5}{a}+\mathrm{3}{b}=\mathrm{1} \\ $$$$\:\:\:{b}=−{a}\:\:\:\:\:\wedge\:\mathrm{5}{a}−\mathrm{3}{a}=\mathrm{1}\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{b}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{3}\right)\left(\mathrm{2}{n}+\mathrm{5}\right)}=\frac{\mathrm{1}/\mathrm{2}}{\mathrm{2}{n}+\mathrm{3}}+\frac{−\mathrm{1}/\mathrm{2}}{\mathrm{2}{n}+\mathrm{5}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{5}}\right) \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}\left\{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{5}}\right)\right\} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{5}}\right) \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}}\right)−\underset{{n}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{5}}\right) \\ $$$$\:\:\:=\left(\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{7}}\right)+\left(\frac{\mathrm{1}}{\mathrm{7}}−\frac{\mathrm{1}}{\mathrm{9}}\right)+\left(\frac{\mathrm{1}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{11}}\right)+\left(\frac{\mathrm{1}}{\mathrm{11}}−\frac{\mathrm{1}}{\mathrm{13}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{13}}=\frac{\mathrm{8}}{\mathrm{65}} \\ $$
Commented by hardmath last updated on 06/Dec/23
thank you professor cool
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{professor}\:\mathrm{cool} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *