Question

Suma de radicali de ordin 2

Answer 1

Răspuns:

a)

Explicație pas cu pas:

In poza

Answer 2

$\displaystyle \dfrac{1}{a+b} = \dfrac{ab}{b^2-a^2}\cdot \Big(\dfrac{1}{a}-\dfrac{1}{b}\Big)\quad (formula)\\ \\\\ S = \sum\limits_{k=1}^{n}\dfrac{1}{(k+1)\sqrt k+k\sqrt{k+1}}\\ \\ S = \sum\limits_{k=1}^{n}\left[\dfrac{k(k+1)\sqrt{k(k+1)}}{k^2(k+1)-(k+1)^2k}\cdot \Big(\dfrac{1}{(k+1)\sqrt k}-\dfrac{1}{k\sqrt{k+1}}\Big)\right]\\ \\ S = \sum\limits_{k=1}^n\left[\dfrac{\sqrt{k(k+1)}}{k-(k+1)}\cdot \Big(\dfrac{1}{(k+1)\sqrt k}-\dfrac{1}{k\sqrt{k+1}}\Big)\right]$

$\displaystyle S = \sum\limits_{k=1}^n\left[\sqrt{k(k+1)}\cdot \Big(\dfrac{1}{k\sqrt{k+1}}-\dfrac{1}{(k+1)\sqrt k}\Big)\right] \\ \\ S = \sum\limits_{k=1}^n\Big(\dfrac{\sqrt k}{k}-\dfrac{\sqrt{k+1}}{k+1}\Big) \\ \\ S=\sum\limits_{k=1}^n\Big(\dfrac{1}{\sqrt k}-\dfrac{1}{\sqrt{k+1}}\Big) \\ \\\\S = \dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+...+\dfrac{1}{\sqrt n}-\dfrac{1}{\sqrt 2}-\dfrac{1}{\sqrt 3}-...-\dfrac{1}{\sqrt n}-\dfrac{1}{\sqrt{n+1}} \\\\\\\Rightarrow \boxed{S = 1 - \dfrac{1}{\sqrt{n+1}}}$