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Question Number 75136 by chess1 last updated on 07/Dec/19

Commented by MJS last updated on 07/Dec/19

$\frac{1}{10}$
Commented by mathmax by abdo last updated on 07/Dec/19
$let decompose F(x)=(1/(4x^2 +24x +35)) 4x^2 +24x +35 =0 ⇒Δ^′ =12^2 −4×35 =144−140 =4 ⇒ x_1 =((−12+2)/4) =−(5/2) and x_2 =((−12−2)/4) =−(7/2) ⇒ F(x)=(1/(4(x+(5/2))(x+(7/2)))) =(1/((2x+5)(2x+7))) =(1/2){ (1/(2x+5))−(1/(2x+7))} let S_n =Σ_(k=0) ^n (1/(4k^2 +24k +35)) ⇒ S_n =(1/2)Σ_(k=0) ^n {(1/(2k+5))−(1/(2k+7))} let v_n =(1/(2n+5)) ⇒v_(n+1) =(1/(2n+7)) ⇒ S_n =(1/2)Σ_(k=0) ^n {v_k −v_(k+1) } =(1/2){v_0 −v_1 +v_1 −v_2 +....+v_n −v_(n+1) } =(1/2){v_0 −v_(n+1) } =(1/2){(1/5)−(1/(2n+7))} ⇒lim_(n→+∞) S_n =(1/(10)) =S$
$l e t d e c o m p o s e F (x) = \frac{1}{4 x^{2} + 24 x + 35}$ $4 x^{2} + 24 x + 35 = 0 \Rightarrow Δ^{^{'}} = 12^{2} - 4 \times 35 = 144 - 140 = 4 \Rightarrow$ $x_{1} = \frac{- 12 + 2}{4} = - \frac{5}{2} a n d x_{2} = \frac{- 12 - 2}{4} = - \frac{7}{2} \Rightarrow$ $F (x) = \frac{1}{4 (x + \frac{5}{2}) (x + \frac{7}{2})} = \frac{1}{(2 x + 5) (2 x + 7)}$ $= \frac{1}{2} {\frac{1}{2 x + 5} - \frac{1}{2 x + 7}} l e t S_{n} = \sum_{k = 0}^{n} \frac{1}{4 k^{2} + 24 k + 35} \Rightarrow$ $S_{n} = \frac{1}{2} \sum_{k = 0}^{n} {\frac{1}{2 k + 5} - \frac{1}{2 k + 7}} l e t v_{n} = \frac{1}{2 n + 5} \Rightarrow v_{n + 1} = \frac{1}{2 n + 7} \Rightarrow$ $S_{n} = \frac{1}{2} \sum_{k = 0}^{n} {v_{k} - v_{k + 1}} = \frac{1}{2} {v_{0} - v_{1} + v_{1} - v_{2} + . . . . + v_{n} - v_{n + 1}}$ $= \frac{1}{2} {v_{0} - v_{n + 1}} = \frac{1}{2} {\frac{1}{5} - \frac{1}{2 n + 7}} \Rightarrow {l i m}_{n \to + \infty} S_{n} = \frac{1}{10} = S$
Commented by chess1 last updated on 08/Dec/19

$thanks$
	Answered by 21042004 last updated on 13/Dec/19

	$0.1$
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