Σ_(k=1) ^n ((k/((4k^2 −1)(2k+3))))=?


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Question Number 155745 by cortano last updated on 04/Oct/21

$\underset{k = 1}{\sum^{n}} (\frac{k}{({4 k}^{2} - 1) (2 k + 3)}) = ?$
	Answered by amin96 last updated on 04/Oct/21
	$(a/(2k−1_(4k^2 +8k+3) ))+(b/(2k+1))_(4k^2 +4k−3) +(c/(2k+3))_(4k^2 −1) 4ak^2 +8ak+3a+4bk^2 +4bk−3b+4ck^2 −c { ((4a+4b+4c=0)),((8a+4b=1)),((3a−3b−c=0 ⇒a−b=(c/3))) :} 2a+b=(1/4) 3a=(c/3)+(1/4) a=(c/9)+(1/(12)) b=a−(c/3)=((c−3c)/9)+(1/(12)) b=(1/(12))−((2c)/9) (c/9)+(1/(12))+(1/(12))−((2c)/9)+c=0 (1/6)−(c/9)+c=0 (1/6)=−((8c)/9) c=−(9/(48))=−(3/(16)) a=(1/(16)) b=(1/8) c=((−3)/(16)) Σ_(k=1) ^n ((1/(16(2k−1)))+(1/(8(2k+1)))−(3/(16(2k+3))))= =(1/(16))Σ_(k=1) ^n ((1/(2k−1))+(2/(2k+1))−(3/(2k+3)))= =(1/(16))(1+(1/3)+(1/5)+…+(1/(2n−1))_(n−1) +(2/3)+(2/5)+(2/7)+…+_(n−1) (2/(2n+1))− −(3/5)−(3/7)−(3/9)−…−(3/(2n+3))) (1/(16))(1+1+(3/5)+(3/7)+(3/9)+…+(3/(2n−1))+(2/(2n+1))−(3/5)−(7/3)−(9/3)−…−(3/(2n+3)))= (1/(16))(2+(2/(2n+1))−(3/(2n+1))−(3/(2n+3)))=(1/(16))(2−(1/(2n+1))−(3/(2n+3)))= (1/(16))(((2(4n^2 +8n+3)−2n−3−6n−3)/((2n+1)(2n+3))))= =(1/(16))(((8n^2 +16n+6−8n−6)/((2n+1)(2n+3))))=(1/(16))(((8n^2 +8n)/((2n+1)(2n+3))))= =((n^2 +n)/(2(2n+1)(2n+3))) MATH.AMIN$
	$\underset{⏟}{\frac{a}{2 k - \underset{4 k^{2} + 8 k + 3}{1}}} + \underset{4 k^{2} + 4 k - 3}{\underset{⏟}{\frac{b}{2 k + 1}}} + \underset{4 k^{2} - 1}{\underset{⏟}{\frac{c}{2 k + 3}}}$ $4 {a k}^{2} + 8 a k + 3 a + 4 {b k}^{2} + 4 b k - 3 b + 4 {c k}^{2} - c$ ${\begin{cases} 4 a + 4 b + 4 c = 0 \\ 8 a + 4 b = 1 \\ 3 a - 3 b - c = 0 \Rightarrow a - b = \frac{c}{3} \end{cases}$ $2 a + b = \frac{1}{4} 3 a = \frac{c}{3} + \frac{1}{4} a = \frac{c}{9} + \frac{1}{12}$ $b = a - \frac{c}{3} = \frac{c - 3 c}{9} + \frac{1}{12}$ $b = \frac{1}{12} - \frac{2 c}{9}$ $\frac{c}{9} + \frac{1}{12} + \frac{1}{12} - \frac{2 c}{9} + c = 0 \frac{1}{6} - \frac{c}{9} + c = 0$ $\frac{1}{6} = - \frac{8 c}{9} c = - \frac{9}{48} = - \frac{3}{16}$ $a = \frac{1}{16} b = \frac{1}{8} c = \frac{- 3}{16}$ $\underset{k = 1}{\sum^{n}} (\frac{1}{16 (2 k - 1)} + \frac{1}{8 (2 k + 1)} - \frac{3}{16 (2 k + 3)}) =$ $= \frac{1}{16} \underset{k = 1}{\sum^{n}} (\frac{1}{2 k - 1} + \frac{2}{2 k + 1} - \frac{3}{2 k + 3}) =$ $= \frac{1}{16} (1 + \underset{n - 1}{\underset{⏟}{\frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{2 n - 1}}} + \underset{n - 1}{\underset{⏟}{\frac{2}{3} + \frac{2}{5} + \frac{2}{7} + \dots +}} \frac{2}{2 n + 1} -$ $- \frac{3}{5} - \frac{3}{7} - \frac{3}{9} - \dots - \frac{3}{2 n + 3})$ $\frac{1}{16} (1 + 1 + \frac{3}{5} + \frac{3}{7} + \frac{3}{9} + \dots + \frac{3}{2 n - 1} + \frac{2}{2 n + 1} - \frac{3}{5} - \frac{7}{3} - \frac{9}{3} - \dots - \frac{3}{2 n + 3}) =$ $\frac{1}{16} (2 + \frac{2}{2 n + 1} - \frac{3}{2 n + 1} - \frac{3}{2 n + 3}) = \frac{1}{16} (2 - \frac{1}{2 n + 1} - \frac{3}{2 n + 3}) =$ $\frac{1}{16} (\frac{2 (4 n^{2} + 8 n + 3) - 2 n - 3 - 6 n - 3}{(2 n + 1) (2 n + 3)}) =$ $= \frac{1}{16} (\frac{8 n^{2} + 16 n + 6 - 8 n - 6}{(2 n + 1) (2 n + 3)}) = \frac{1}{16} (\frac{8 n^{2} + 8 n}{(2 n + 1) (2 n + 3)}) =$ $= \frac{n^{2} + n}{2 (2 n + 1) (2 n + 3)}$ $M A T H . A M I N$
	Commented by Tawa11 last updated on 04/Oct/21

	$Weldone sir$
	Commented by cortano last updated on 05/Oct/21

	$nice$
	Answered by TheSupreme last updated on 04/Oct/21
	$(A/(2k−1))+(B/(2k+1))+(C/(2k+3))=((A(2k+1)(2k+3)+B(2k−1)(2k+3)+C(4k^2 −1))/D) { ((A+B+C=0)),((2A+B=(1/4))),((3A−3B−C=0)) :} A=−C−B −2C−2B+B=(1/4)→ B=2C−(1/4) −3C−3(2C−(1/4))−3(2C−(1/4))−C=0 −3C−6C+(3/4)−6C+(3/4)−C=0 −15C=−(6/4) → C=(1/(10)) B=(1/5)−(1/4)=−(1/(20)) A=−(1/(20)) Σ=(1/(10))[−(1/2)(1/(2k+1))−(1/2)(1/(2k−1))+(1/(2k+3))] =(1/(10))[−(1/2)( (1/1)+(1/3)+(1/5)+...)−(1/2)((1/3)+(1/5)+...)+(1/5)] =(1/(10))[−(1/2)−(1/2)((1/3))]=−(1/(15))$
	$\frac{A}{2 k - 1} + \frac{B}{2 k + 1} + \frac{C}{2 k + 3} = \frac{A (2 k + 1) (2 k + 3) + B (2 k - 1) (2 k + 3) + C (4 k^{2} - 1)}{D}$ ${\begin{cases} A + B + C = 0 \\ 2 A + B = \frac{1}{4} \\ 3 A - 3 B - C = 0 \end{cases}$ $A = - C - B$ $- 2 C - 2 B + B = \frac{1}{4} \to B = 2 C - \frac{1}{4}$ $- 3 C - 3 (2 C - \frac{1}{4}) - 3 (2 C - \frac{1}{4}) - C = 0$ $- 3 C - 6 C + \frac{3}{4} - 6 C + \frac{3}{4} - C = 0$ $- 15 C = - \frac{6}{4} \to C = \frac{1}{10}$ $B = \frac{1}{5} - \frac{1}{4} = - \frac{1}{20}$ $A = - \frac{1}{20}$ $Σ = \frac{1}{10} [- \frac{1}{2} \frac{1}{2 k + 1} - \frac{1}{2} \frac{1}{2 k - 1} + \frac{1}{2 k + 3}]$ $= \frac{1}{10} [- \frac{1}{2} (\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + . . .) - \frac{1}{2} (\frac{1}{3} + \frac{1}{5} + . . .) + \frac{1}{5}]$ $= \frac{1}{10} [- \frac{1}{2} - \frac{1}{2} (\frac{1}{3})] = - \frac{1}{15}$
	Commented by amin96 last updated on 04/Oct/21

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