Question

Va rog sa ma ajutati cu aceasta problema.

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

Răspuns:

$S =1- (k+2)\cdot 3^{-k-1}$

Rezolvare:

$S = \dfrac{1}{3}+\dfrac{3}{3^2}+\dfrac{5}{3^3}+...+\dfrac{2k+1}{3^{k+1}}\\ \\ S = 1\cdot 3^{-1}+3\cdot 3^{-2}+5\cdot 3^{-3}+...+(2k+1)\cdot 3^{-(k+1)}\\ \\ S(x) = 1\cdot x^{-1}+3\cdot x^{-2}+5\cdot x^{-3}+...+(2k+1)\cdot x^{-(k+1)}\\ \\ S(x) + x^{-1}+x^{-2}+...+x^{-(k+1)} = 2x^{-1}+4x^{-2}+...+2(k+1)x^{-(k+1)}\\ \\ S(x)+x^{-1}\cdot \dfrac{x^{-k-1}-1}{x^{-1}-1} = 2\Big(x^{-1}+ 2x^{-2}+3x^{-3}+...+(k+1)x^{-(k-1)}\Big)$

$x^{-1}S(x)+\dfrac{x^{-k-2}-x^{-1}}{1-x} = -2\Big(-x^{-2}-2x^{-3}-3x^{-4}-...-(k+1)x^{-(k+2)}\Big)\\ \\\\\text{Notez: }\, \sigma(x) = -x^{-2}-2x^{-3}-3x^{-4}-...-(k+1)x^{-(k+2)}\\ \\ \sigma(x) = \sum\limits_{i=1}^{k+1}(-ix^{-i-1})\\ \\ \int\sigma(x)\, dx =\sum\limits_{i=1}^{k+1}(x^{-i})+C\\ \\ \int\sigma(x)\, dx = x^{-1}+x^{-2}+...+x^{-(k+1)}+C\\ \\ \int \sigma(x)\, dx = x^{-1}\cdot \dfrac{x^{-k-1}-1}{x^{-1}-1}+C\Big|'$

$\sigma(x) = \Big(\dfrac{x^{-k-2}-x^{-1}}{x^{-1}-1}\Big)' \\ \\ \sigma(x) =\dfrac{\Big[(-k-2)x^{-k-3}+x^{-2}\Big](x^{-1}-1)+x^{-2}(x^{-k-2}-x^{-1})}{(x^{-1}-1)^2}$

$x^{-1}S(x)+\dfrac{x^{-k-2}-x^{-1}}{1-x} = -2\sigma(x)\\ \\\\\text{Fac }\, x = 3: \\ \\ \dfrac{S(3)}{3}+\dfrac{3^{-k-2}-3^{-1}}{1-3} = -2\sigma(3)\\ \\\\\sigma(3) =\dfrac{\Big[(-k-2)3^{-k-3}+3^{-2}\Big](3^{-1}-1)+3^{-2}(3^{-k-2}-3^{-1})}{(3^{-1}-1)^2}\\ \\ \sigma(3) = \dfrac{9}{4}\cdot \Bigg(\Big((-k-2)3^{-k-3}+3^{-2}\Big)(3^{-1}-1)+3^{-2}(3^{-k-2}-3^{-1})\Bigg)$

$\sigma(3) = \dfrac{1}{4}\cdot \Big((k+2)3^{-k-1}-1)\cdot \dfrac{2}{3} + 3^{-k-2}-3^{-1}\Big)\\ \\ 12\sigma(3) = 2(k+2)3^{-k-1}-2+3^{-k-1}-1$

$\dfrac{S(3)}{3}+\dfrac{3^{-k-2}-3^{-1}}{1-3} = -2\sigma(3) \\ \\ 2S(3)-3^{-k-1}+1 = -12\sigma(3)\\ \\ 2S(3) = -\Big[2(k+2)3^{-k-1}-2+3^{-k-1}-1\Big]+3^{-k-1}-1\\ \\ 2S(3) = -2(k+2)3^{-k-1}+2 \\ \\ S(3) =-(k+2)\cdot 3^{-k-1}+1\\\\\\\Rightarrow \boxed{S =1 - (k+2)\cdot 3^{-k-1}}$