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Question Number 130472 by shaker last updated on 26/Jan/21

	Answered by mathmax by abdo last updated on 26/Jan/21
	$let w(x) =Σ_(k=1) ^n (2k−1)x^k ⇒w(x)=2Σ_(k=1) ^n kx^k −Σ_(k=1) ^n x^k we have Σ_(k=0) ^n x^k =((x^(n+1) −1)/(x−1)) (x≠1) ⇒Σ_(k=1) ^n kx^(k−1) =((nx^(n+1) −(n+1)x^n +1)/((x−1)^2 )) ⇒Σ_(k=1) ^n kx^k =(x/((x−1)^2 ))(nx^(n+1) −(n+1)x^n +1) ⇒ w(x)=((2x)/((x−1)^2 ))(nx^(n+1) −(n+1)x^n +1)−((1/(1−x))−1) =((2x(nx^(n+1) −(n+1)x^n +1))/((x−1)^2 ))−((x(1−x))/((1−x)^2 )) =((2x(nx^(n+1) −(n+1)x^n +1)+x^2 −x)/((x−1)^2 )) ⇒ f(x)=(((√(2nx^(n+2) −2(n+1)x^(n+1) +x^2 +x))×(1/(x−1))−n)/((x−1))) =(((√(2nx^(n+2) −2(n+1)x^(n+1) +x^2 +x))−n(x−1))/((x−1)^2 )) we do the changement x−1=t (t→0) ⇒f(x)=f(t+1) =(((√(2n(t+1)^(n+2) −2(n+1)(t+1)^(n+1) +(t+1)^2 +t))−nt)/t^2 ) we have (1+t)^(n+2) ∼1+(n+2)t +(((n+2)(n+1))/2)t^2 (1+t)^(n+1) ∼1+(n+1)t+(((n+1)n)/2)t^2 (t+1)^2 +t ∼1 ⇒ 2n(t+1)^(n+2) −2(n+1)(t+1)^(n+1) +(t+1)^2 +t ∼2n{1+(n+2)t+(((n+2)(n+1))/2)t^2 }−2(n+1)(1+(n+1)t+(((n+1)n)/2)t^2 )+1 ....be continued...$
	$let w (x) = \sum_{k = 1}^{n} (2 k - 1) x^{k} \Rightarrow w (x) = 2 \sum_{k = 1}^{n} {kx}^{k} - \sum_{k = 1}^{n} x^{k}$ $we have \sum_{k = 0}^{n} x^{k} = \frac{x^{n + 1} - 1}{x - 1} (x \neq 1) \Rightarrow \sum_{k = 1}^{n} {kx}^{k - 1} = \frac{{nx}^{n + 1} - (n + 1) x^{n} + 1}{{(x - 1)}^{2}}$ $\Rightarrow \sum_{k = 1}^{n} {kx}^{k} = \frac{x}{{(x - 1)}^{2}} ({nx}^{n + 1} - (n + 1) x^{n} + 1) \Rightarrow$ $w (x) = \frac{2 x}{{(x - 1)}^{2}} ({nx}^{n + 1} - (n + 1) x^{n} + 1) - (\frac{1}{1 - x} - 1)$ $= \frac{2 x ({nx}^{n + 1} - (n + 1) x^{n} + 1)}{{(x - 1)}^{2}} - \frac{x (1 - x)}{{(1 - x)}^{2}}$ $= \frac{2 x ({nx}^{n + 1} - (n + 1) x^{n} + 1) + x^{2} - x}{{(x - 1)}^{2}} \Rightarrow$ $f (x) = \frac{\sqrt{{2 nx}^{n + 2} - 2 (n + 1) x^{n + 1} + x^{2} + x} \times \frac{1}{x - 1} - n}{(x - 1)}$ $= \frac{\sqrt{{2 nx}^{n + 2} - 2 (n + 1) x^{n + 1} + x^{2} + x} - n (x - 1)}{{(x - 1)}^{2}} we do the changement$ $x - 1 = t (t \to 0) \Rightarrow f (x) = f (t + 1)$ $= \frac{\sqrt{2 n {(t + 1)}^{n + 2} - 2 (n + 1) {(t + 1)}^{n + 1} + {(t + 1)}^{2} + t} - nt}{t^{2}}$ $we have {(1 + t)}^{n + 2} \sim 1 + (n + 2) t + \frac{(n + 2) (n + 1)}{2} t^{2}$ ${(1 + t)}^{n + 1} \sim 1 + (n + 1) t + \frac{(n + 1) n}{2} t^{2}$ ${(t + 1)}^{2} + t \sim 1 \Rightarrow$ $2 n {(t + 1)}^{n + 2} - 2 (n + 1) {(t + 1)}^{n + 1} + {(t + 1)}^{2} + t$ $\sim 2 n {1 + (n + 2) t + \frac{(n + 2) (n + 1)}{2} t^{2}} - 2 (n + 1) (1 + (n + 1) t + \frac{(n + 1) n}{2} t^{2}) + 1$ $. . . . be continued . . .$
	Answered by mathmax by abdo last updated on 26/Jan/21
	$in this case it better to use[hospital theorem let u(x)=(√(x+3x^2 +5x^3 +....+(2n−1)x^n ))−n and v(x)=x−1 u^′ (x)=((1+6x+15x^2 +....+n(2n−1)x^(n−1) )/(2(√(x+3x^2 +5x^3 +....+(2n−1)x^n )))) ⇒ lim_(x→1) u^′ (x)=((1+6+15+....+n(2n−1))/(2(√(1+3+5+....+2n−1)))) v^′ (x)=1 ⇒lim_(x→1) v^′ (x)=1 also Σ_(k=1) ^n k(2k−1) =2Σ_(k=1) ^n k^2 −Σ_(k=1) ^n k =2.((n(n+1)(2n+1))/6)−((n(n+1))/2) =((n(n+1)(2n+1))/3)−((n(n+1))/2)=n(n+1){((2n+1)/3)−(1/2)} =n(n+1)(((4n+2−3)/6)) =((n(n+1)(4n−1))/6) Σ_(k=1) ^n (2k−1) =2Σ_(k=1) ^n k−Σ_(k=1) ^n k =2((n(n+1))/2)−((n(n+1))/2) =((n(n+1))/2) ⇒lim_(x→1) (...) =(((n(n+1)(4n−1))/6)/(2(√((n(n+1))/2)))) =((n(n+1)(4n−1))/(6(√2)(√(n(n+1))))) =(((4n−1)(√(n(n+1))))/(6(√2)))$
	$in this case it better to use [hospital theorem let$ $u (x) = \sqrt{x + {3 x}^{2} + {5 x}^{3} + . . . . + (2 n - 1) x^{n}} - n and v (x) = x - 1$ $u^{^{'}} (x) = \frac{1 + 6 x + {15 x}^{2} + . . . . + n (2 n - 1) x^{n - 1}}{2 \sqrt{x + {3 x}^{2} + {5 x}^{3} + . . . . + (2 n - 1) x^{n}}} \Rightarrow$ $\lim_{x \to 1} u^{^{'}} (x) = \frac{1 + 6 + 15 + . . . . + n (2 n - 1)}{2 \sqrt{1 + 3 + 5 + . . . . + 2 n - 1}}$ $v^{^{'}} (x) = 1 \Rightarrow \lim_{x \to 1} v^{^{'}} (x) = 1 also$ $\sum_{k = 1}^{n} k (2 k - 1) = 2 \sum_{k = 1}^{n} k^{2} - \sum_{k = 1}^{n} k = 2 . \frac{n (n + 1) (2 n + 1)}{6} - \frac{n (n + 1)}{2}$ $= \frac{n (n + 1) (2 n + 1)}{3} - \frac{n (n + 1)}{2} = n (n + 1) {\frac{2 n + 1}{3} - \frac{1}{2}}$ $= n (n + 1) (\frac{4 n + 2 - 3}{6}) = \frac{n (n + 1) (4 n - 1)}{6}$ $\sum_{k = 1}^{n} (2 k - 1) = 2 \sum_{k = 1}^{n} k - \sum_{k = 1}^{n} k = 2 \frac{n (n + 1)}{2} - \frac{n (n + 1)}{2}$ $= \frac{n (n + 1)}{2} \Rightarrow \lim_{x \to 1} (. . .)$ $= \frac{\frac{n (n + 1) (4 n - 1)}{6}}{2 \sqrt{\frac{n (n + 1)}{2}}} = \frac{n (n + 1) (4 n - 1)}{6 \sqrt{2} \sqrt{n (n + 1)}}$ $= \frac{(4 n - 1) \sqrt{n (n + 1)}}{6 \sqrt{2}}$
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