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Find-1-n-1-16-16n-2-8n-3-2-n-1-1-n-2n-3-

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Question Number 202468 by hardmath last updated on 27/Dec/23
Find: 1. Σ_(n=1) ^( ∞) ((16)/(16n^2 − 8n − 3)) = ? 2. Σ_(n=1) ^( ∞) (((−1)^n )/(2n^3 )) = ?
$$\mathrm{Find}: \\ $$$$\mathrm{1}.\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\:\infty} {\sum}}\:\frac{\mathrm{16}}{\mathrm{16n}^{\mathrm{2}} \:−\:\mathrm{8n}\:−\:\mathrm{3}}\:=\:?\:\:\:\:\:\mathrm{2}.\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\:\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} }{\mathrm{2n}^{\mathrm{3}} }\:=\:? \\ $$
Answered by Rasheed.Sindhi last updated on 27/Dec/23
1. Σ_(n=1) ^(∞) ((16)/(16n^2 −8n−3)) ((16)/(16n^2 −8n−3))=((16)/((4n−3)(4n+1)))=(a/(4n−3))+(b/(4n+1)) 16=a(4n+1)+b(4n−3) n=−1/4: 16=−4b⇒b=−4 n=(3/4) : 16=4a⇒a=4 Σ_(n=1) ^(∞) ((16)/(16n^2 −8n−3))=Σ_(n=1) ^(∞) ((4/(4n−3))−(4/(4n+1))) determinant ((t_1 ,((4/1)−(4/5))),(t_2 ,((4/5)−(4/9))),(t_3 ,((4/9)−(4/(13)))),((...),(...)),(t_(n−1) ,((4/(4n−7))−(4/(4n−3)))),(t_n ,((4/(4n−3))−(4/(4n+1)))),( ,(Σ_(n=1) ^(n=n) (4/(4n−3))−(4/(4n+1))=4−(4/(4n+1))))) Σ_(n=1) ^(∞) ( 4−(4/(4n+1)))=lim_(x→∞) ( 4−(4/(4n+1)))=4−0=4
$$\mathrm{1}.\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\Sigma}}\:\:\frac{\mathrm{16}}{\mathrm{16}{n}^{\mathrm{2}} −\mathrm{8}{n}−\mathrm{3}} \\ $$$$\frac{\mathrm{16}}{\mathrm{16}{n}^{\mathrm{2}} −\mathrm{8}{n}−\mathrm{3}}=\frac{\mathrm{16}}{\left(\mathrm{4}{n}−\mathrm{3}\right)\left(\mathrm{4}{n}+\mathrm{1}\right)}=\frac{{a}}{\mathrm{4}{n}−\mathrm{3}}+\frac{{b}}{\mathrm{4}{n}+\mathrm{1}} \\ $$$$\mathrm{16}={a}\left(\mathrm{4}{n}+\mathrm{1}\right)+{b}\left(\mathrm{4}{n}−\mathrm{3}\right) \\ $$$${n}=−\mathrm{1}/\mathrm{4}:\:\mathrm{16}=−\mathrm{4}{b}\Rightarrow{b}=−\mathrm{4} \\ $$$${n}=\frac{\mathrm{3}}{\mathrm{4}}\::\:\mathrm{16}=\mathrm{4}{a}\Rightarrow{a}=\mathrm{4} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\Sigma}}\:\:\frac{\mathrm{16}}{\mathrm{16}{n}^{\mathrm{2}} −\mathrm{8}{n}−\mathrm{3}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\Sigma}}\:\left(\frac{\mathrm{4}}{\mathrm{4}{n}−\mathrm{3}}−\frac{\mathrm{4}}{\mathrm{4}{n}+\mathrm{1}}\right) \\ $$$$\begin{array}{|c|c|c|c|c|c|c|}{{t}_{\mathrm{1}} }&\hline{\frac{\mathrm{4}}{\mathrm{1}}−\cancel{\frac{\mathrm{4}}{\mathrm{5}}}}\\{{t}_{\mathrm{2}} }&\hline{\cancel{\frac{\mathrm{4}}{\mathrm{5}}}−\cancel{\frac{\mathrm{4}}{\mathrm{9}}}}\\{{t}_{\mathrm{3}} }&\hline{\cancel{\frac{\mathrm{4}}{\mathrm{9}}}−\cancel{\frac{\mathrm{4}}{\mathrm{13}}}}\\{…}&\hline{…}\\{{t}_{{n}−\mathrm{1}} }&\hline{\cancel{\frac{\mathrm{4}}{\mathrm{4}{n}−\mathrm{7}}}−\cancel{\frac{\mathrm{4}}{\mathrm{4}{n}−\mathrm{3}}}}\\{{t}_{{n}} }&\hline{\cancel{\frac{\mathrm{4}}{\mathrm{4}{n}−\mathrm{3}}}−\frac{\mathrm{4}}{\mathrm{4}{n}+\mathrm{1}}}\\{\:}&\hline{\underset{{n}=\mathrm{1}} {\overset{{n}={n}} {\sum}}\:\frac{\mathrm{4}}{\mathrm{4}{n}−\mathrm{3}}−\frac{\mathrm{4}}{\mathrm{4}{n}+\mathrm{1}}=\mathrm{4}−\frac{\mathrm{4}}{\mathrm{4}{n}+\mathrm{1}}}\\\hline\end{array}\: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\Sigma}}\:\:\left(\:\mathrm{4}−\frac{\mathrm{4}}{\mathrm{4}{n}+\mathrm{1}}\right)=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\:\mathrm{4}−\frac{\mathrm{4}}{\mathrm{4}{n}+\mathrm{1}}\right)=\mathrm{4}−\mathrm{0}=\mathrm{4} \\ $$
Commented by hardmath last updated on 30/Dec/23
thankyou dear professor cool
$$\mathrm{thankyou}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{cool} \\ $$

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