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Question Number 155745 by cortano last updated on 04/Oct/21
Σ_(k=1) ^n ((k/((4k^2 −1)(2k+3))))=?
$$\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\frac{\mathrm{k}}{\left(\mathrm{4k}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{2k}+\mathrm{3}\right)}\right)=? \\ $$
Answered by amin96 last updated on 04/Oct/21
(a/(2k−1_(4k^2 +8k+3) ))+(b/(2k+1))_(4k^2 +4k−3) +(c/(2k+3))_(4k^2 −1) 4ak^2 +8ak+3a+4bk^2 +4bk−3b+4ck^2 −c { ((4a+4b+4c=0)),((8a+4b=1)),((3a−3b−c=0 ⇒a−b=(c/3))) :} 2a+b=(1/4) 3a=(c/3)+(1/4) a=(c/9)+(1/(12)) b=a−(c/3)=((c−3c)/9)+(1/(12)) b=(1/(12))−((2c)/9) (c/9)+(1/(12))+(1/(12))−((2c)/9)+c=0 (1/6)−(c/9)+c=0 (1/6)=−((8c)/9) c=−(9/(48))=−(3/(16)) a=(1/(16)) b=(1/8) c=((−3)/(16)) Σ_(k=1) ^n ((1/(16(2k−1)))+(1/(8(2k+1)))−(3/(16(2k+3))))= =(1/(16))Σ_(k=1) ^n ((1/(2k−1))+(2/(2k+1))−(3/(2k+3)))= =(1/(16))(1+(1/3)+(1/5)+…+(1/(2n−1))_(n−1) +(2/3)+(2/5)+(2/7)+…+_(n−1) (2/(2n+1))− −(3/5)−(3/7)−(3/9)−…−(3/(2n+3))) (1/(16))(1+1+(3/5)+(3/7)+(3/9)+…+(3/(2n−1))+(2/(2n+1))−(3/5)−(7/3)−(9/3)−…−(3/(2n+3)))= (1/(16))(2+(2/(2n+1))−(3/(2n+1))−(3/(2n+3)))=(1/(16))(2−(1/(2n+1))−(3/(2n+3)))= (1/(16))(((2(4n^2 +8n+3)−2n−3−6n−3)/((2n+1)(2n+3))))= =(1/(16))(((8n^2 +16n+6−8n−6)/((2n+1)(2n+3))))=(1/(16))(((8n^2 +8n)/((2n+1)(2n+3))))= =((n^2 +n)/(2(2n+1)(2n+3))) MATH.AMIN
$$\underbrace{\frac{{a}}{\mathrm{2}{k}−\underset{\mathrm{4}{k}^{\mathrm{2}} +\mathrm{8}{k}+\mathrm{3}} {\mathrm{1}}}}+\underset{\mathrm{4}{k}^{\mathrm{2}} +\mathrm{4}{k}−\mathrm{3}} {\underbrace{\frac{{b}}{\mathrm{2}{k}+\mathrm{1}}}}+\underset{\mathrm{4}{k}^{\mathrm{2}} −\mathrm{1}} {\underbrace{\frac{{c}}{\mathrm{2}{k}+\mathrm{3}}}} \\ $$$$\mathrm{4}{ak}^{\mathrm{2}} +\mathrm{8}{ak}+\mathrm{3}{a}+\mathrm{4}{bk}^{\mathrm{2}} +\mathrm{4}{bk}−\mathrm{3}{b}+\mathrm{4}{ck}^{\mathrm{2}} −{c} \\ $$$$\begin{cases}{\mathrm{4}{a}+\mathrm{4}{b}+\mathrm{4}{c}=\mathrm{0}}\\{\mathrm{8}{a}+\mathrm{4}{b}=\mathrm{1}}\\{\mathrm{3}{a}−\mathrm{3}{b}−{c}=\mathrm{0}\:\:\Rightarrow{a}−{b}=\frac{{c}}{\mathrm{3}}}\end{cases}\: \\ $$$$\mathrm{2}{a}+{b}=\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\mathrm{3}{a}=\frac{{c}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:{a}=\frac{{c}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{12}} \\ $$$${b}={a}−\frac{{c}}{\mathrm{3}}=\frac{{c}−\mathrm{3}{c}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{12}} \\ $$$${b}=\frac{\mathrm{1}}{\mathrm{12}}−\frac{\mathrm{2}{c}}{\mathrm{9}} \\ $$$$\frac{{c}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{12}}−\frac{\mathrm{2}{c}}{\mathrm{9}}+{c}=\mathrm{0}\:\:\:\:\frac{\mathrm{1}}{\mathrm{6}}−\frac{{c}}{\mathrm{9}}+{c}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}=−\frac{\mathrm{8}{c}}{\mathrm{9}}\:\:\:\:\:\:{c}=−\frac{\mathrm{9}}{\mathrm{48}}=−\frac{\mathrm{3}}{\mathrm{16}} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{16}}\:\:{b}=\frac{\mathrm{1}}{\mathrm{8}}\:\:\:{c}=\frac{−\mathrm{3}}{\mathrm{16}} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{16}\left(\mathrm{2}{k}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{2}{k}+\mathrm{1}\right)}−\frac{\mathrm{3}}{\mathrm{16}\left(\mathrm{2}{k}+\mathrm{3}\right)}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}+\frac{\mathrm{2}}{\mathrm{2}{k}+\mathrm{1}}−\frac{\mathrm{3}}{\mathrm{2}{k}+\mathrm{3}}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{1}+\underset{{n}−\mathrm{1}} {\underbrace{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+\ldots+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}}}+\underset{{n}−\mathrm{1}} {\underbrace{\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{5}}+\frac{\mathrm{2}}{\mathrm{7}}+\ldots+}}\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}−\right. \\ $$$$\left.−\frac{\mathrm{3}}{\mathrm{5}}−\frac{\mathrm{3}}{\mathrm{7}}−\frac{\mathrm{3}}{\mathrm{9}}−\ldots−\frac{\mathrm{3}}{\mathrm{2}{n}+\mathrm{3}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{1}+\mathrm{1}+\frac{\mathrm{3}}{\mathrm{5}}+\frac{\mathrm{3}}{\mathrm{7}}+\frac{\mathrm{3}}{\mathrm{9}}+\ldots+\frac{\mathrm{3}}{\mathrm{2}{n}−\mathrm{1}}+\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}−\frac{\mathrm{3}}{\mathrm{5}}−\frac{\mathrm{7}}{\mathrm{3}}−\frac{\mathrm{9}}{\mathrm{3}}−\ldots−\frac{\mathrm{3}}{\mathrm{2}{n}+\mathrm{3}}\right)= \\ $$$$\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}−\frac{\mathrm{3}}{\mathrm{2}{n}+\mathrm{1}}−\frac{\mathrm{3}}{\mathrm{2}{n}+\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}−\frac{\mathrm{3}}{\mathrm{2}{n}+\mathrm{3}}\right)= \\ $$$$\frac{\mathrm{1}}{\mathrm{16}}\left(\frac{\mathrm{2}\left(\mathrm{4}{n}^{\mathrm{2}} +\mathrm{8}{n}+\mathrm{3}\right)−\mathrm{2}{n}−\mathrm{3}−\mathrm{6}{n}−\mathrm{3}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\left(\frac{\mathrm{8}{n}^{\mathrm{2}} +\mathrm{16}{n}+\mathrm{6}−\mathrm{8}{n}−\mathrm{6}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)}\right)=\frac{\mathrm{1}}{\mathrm{16}}\left(\frac{\mathrm{8}{n}^{\mathrm{2}} +\mathrm{8}{n}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)}\right)= \\ $$$$=\frac{{n}^{\mathrm{2}} +{n}}{\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$$${MATH}.{AMIN}\: \\ $$$$ \\ $$
Commented by Tawa11 last updated on 04/Oct/21
Weldone sir
$$\mathrm{Weldone}\:\mathrm{sir} \\ $$
Commented by cortano last updated on 05/Oct/21
nice
$$\mathrm{nice} \\ $$$$ \\ $$
Answered by TheSupreme last updated on 04/Oct/21
(A/(2k−1))+(B/(2k+1))+(C/(2k+3))=((A(2k+1)(2k+3)+B(2k−1)(2k+3)+C(4k^2 −1))/D) { ((A+B+C=0)),((2A+B=(1/4))),((3A−3B−C=0)) :} A=−C−B −2C−2B+B=(1/4)→ B=2C−(1/4) −3C−3(2C−(1/4))−3(2C−(1/4))−C=0 −3C−6C+(3/4)−6C+(3/4)−C=0 −15C=−(6/4) → C=(1/(10)) B=(1/5)−(1/4)=−(1/(20)) A=−(1/(20)) Σ=(1/(10))[−(1/2)(1/(2k+1))−(1/2)(1/(2k−1))+(1/(2k+3))] =(1/(10))[−(1/2)( (1/1)+(1/3)+(1/5)+...)−(1/2)((1/3)+(1/5)+...)+(1/5)] =(1/(10))[−(1/2)−(1/2)((1/3))]=−(1/(15))
$$\frac{{A}}{\mathrm{2}{k}−\mathrm{1}}+\frac{{B}}{\mathrm{2}{k}+\mathrm{1}}+\frac{{C}}{\mathrm{2}{k}+\mathrm{3}}=\frac{{A}\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{3}\right)+{B}\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{3}\right)+{C}\left(\mathrm{4}{k}^{\mathrm{2}} −\mathrm{1}\right)}{{D}} \\ $$$$\begin{cases}{{A}+{B}+{C}=\mathrm{0}}\\{\mathrm{2}{A}+{B}=\frac{\mathrm{1}}{\mathrm{4}}}\\{\mathrm{3}{A}−\mathrm{3}{B}−{C}=\mathrm{0}}\end{cases} \\ $$$${A}=−{C}−{B} \\ $$$$−\mathrm{2}{C}−\mathrm{2}{B}+{B}=\frac{\mathrm{1}}{\mathrm{4}}\rightarrow\:{B}=\mathrm{2}{C}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$−\mathrm{3}{C}−\mathrm{3}\left(\mathrm{2}{C}−\frac{\mathrm{1}}{\mathrm{4}}\right)−\mathrm{3}\left(\mathrm{2}{C}−\frac{\mathrm{1}}{\mathrm{4}}\right)−{C}=\mathrm{0} \\ $$$$−\mathrm{3}{C}−\mathrm{6}{C}+\frac{\mathrm{3}}{\mathrm{4}}−\mathrm{6}{C}+\frac{\mathrm{3}}{\mathrm{4}}−{C}=\mathrm{0} \\ $$$$−\mathrm{15}{C}=−\frac{\mathrm{6}}{\mathrm{4}}\:\rightarrow\:{C}=\frac{\mathrm{1}}{\mathrm{10}} \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{4}}=−\frac{\mathrm{1}}{\mathrm{20}} \\ $$$${A}=−\frac{\mathrm{1}}{\mathrm{20}} \\ $$$$\Sigma=\frac{\mathrm{1}}{\mathrm{10}}\left[−\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{3}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}\left[−\frac{\mathrm{1}}{\mathrm{2}}\left(\:\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+…\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+…\right)+\frac{\mathrm{1}}{\mathrm{5}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}\left[−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right]=−\frac{\mathrm{1}}{\mathrm{15}} \\ $$$$ \\ $$
Commented by amin96 last updated on 04/Oct/21
the answer should be n
$${the}\:{answer}\:{should}\:{be}\:{n} \\ $$

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