Question

ma poate ajuta cineva va rog ​

Answer 1

$\displaystyle \int ( 1+x+x^2+x^3+...+x^{n-1})\, dx = \\ \\ = x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+...+\dfrac{x^n}{n}+C \\ \\ \int \dfrac{x^n - 1}{x-1}\, dx = x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+...+\dfrac{x^n}{n}+C\\ \\ \int_{0}^{\frac{1}{2}} \dfrac{x^n - 1}{x-1}\, dx = \dfrac{1}{1\cdot 2} + \dfrac{1}{2\cdot 2^2} +\dfrac{1}{3\cdot 2^3}+ ...+\dfrac{1}{n\cdot 2^n} \\ \\ \\ \lim\limits_{n\to \infty} \Bigg(\int_{0}^{\frac{1}{2}} \dfrac{x^n - 1}{x-1}\, dx\Bigg) =\lim\limits_{n\to \infty} \Bigg(\dfrac{1}{1\cdot 2} + \dfrac{1}{2\cdot 2^2} +\dfrac{1}{3\cdot 2^3}+...+\dfrac{1}{n\cdot 2^n} \Bigg)\\ \\ n\to \infty\\ \\0 \leq x\leq \dfrac{1}{2} \Rightarrow x^n \to 0 \\ \\ \Rightarrow \int_{0}^{\frac{1}{2}} \dfrac{-1}{x-1}\, dx = l \\ \\ \Rightarrow -\ln|x-1|\Big|_0^{\frac{1}{2}} = l \\ \\ \Rightarrow -\ln\Big(1-\dfrac{1}{2} \Big) = l \\ \\ \Rightarrow l = \ln(2)$