Question

Se consideră funcția

$\it f:\mathbb{R} \longrightarrow \mathbb{R},\ f(x) = e^x+e^{-x}.\\ \\ \\ Fie\ g(x)=f'(x)+f''(x).$

Calculați suma

$\it S = g(0)+g(1)+g(2)+\ ...\ +g(2019)$

Mulțumesc!!!

Answer 1

$f:\mathbb{R}\to \mathbb{R}. \quad f(x) = e^x+e^{-x}\\ \\ g(x) = f'(x)+f''(x) \\ g(x) = (e^x-e^{-x})+(e^x+e^{-x}) = 2e^x\\ \\ S = g(0)+g(1)+g(2)+...+g(2019) \\ \\S = \displaystyle \sum\limits_{k=0}^{2019}g(k) = \sum\limits_{k=0}^{2019}(2e^{k}) = 2\sum\limits_{k=0}^{2019}e^k = \\ \\ = 2\cdot(e^0+e^1+e^2+...+e^{2019}) =\\ \\ = 2\cdot \dfrac{b_1(q^{n}-1)}{q-1}=2\cdot \dfrac{e^0\cdot (e^{2020}-1)}{e-1} =\boxed{\dfrac{2(e^{2020}-1)}{e-1}}$

$\text{Avem progresie geometrica cu: } b_1 = e^0,\quad q = e,\quad n = 2020$