Respostas

2013-11-05T03:20:02-02:00
Para facilitar o cálculo vou dividir a fórmula assim:

E=E_1^{E_2}

Onde,

E_1=\left[\dfrac{2002}{2\cdot6}+\dfrac{2002}{6\cdot10}+\dfrac{2002}{10\cdot14}+\ldots + \dfrac{2002}{1998\cdot 2002}\right]

E_2=\dfrac{1}{\sqrt1+\sqrt2}+\dfrac{1}{\sqrt2+\sqrt3}+\ldots +\dfrac{1}{\sqrt{99}+\sqrt{100}}

Primeiramente vamos trabalhar com E_1. Assim:

E_1=\dfrac{2002}{2\cdot6}+\dfrac{2002}{6\cdot10}+\dfrac{2002}{10\cdot14}+\ldots + \dfrac{2002}{1998\cdot 2002}

E_1=2002.\left[\dfrac{1}{2.1\cdot2.3}+\dfrac{1}{2.3\cdot2.5}+\dfrac{1}{2.5\cdot2.7}+\ldots + \dfrac{1}{2.999\cdot 2.1001}\right]

E_1=\dfrac{2002}{2.2}.\left[\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\ldots + \dfrac{1}{999\cdot1001}\right]

E_1=\dfrac{1001}{2}.\left[\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\ldots + \dfrac{1}{999\cdot1001}\right]

Agora, vamos dividir novamente. Assim:

E_1=\frac{1001}{2}.E_{3}

Onde,

E_3=\left[\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\ldots + \dfrac{1}{999\cdot1001}\right]

Podemos perceber que esta soma tem a forma:

S_{n}=\left[\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\ldots + \dfrac{1}{(2n+1)\cdot(2n+3)}\right]

Onde n vale:

2n+1=999

2n=999-1

2n=998

n=\frac{998}{2}

n=499

Agora vamos encontrar uma fórmula geral para S_{499}. Não vou provar porque senão fica muito longo, mas temos que:

E_3=S_{n}=\frac{n}{2n+1}

\left[\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\ldots + \dfrac{1}{(2n+1)\cdot(2n+3)}\right]=\frac{n}{2n+1}

Substituindo o valor de n=499 em S_{n} temos:

S_{n}=\frac{n}{2n+1}

S_{499}=\frac{499}{2.499+1}

S_{499}=\frac{499}{999}

E_3=\frac{499}{999}

Voltando para,

E_1=\frac{1001}{2}.E_{3}

E_1=\frac{1001}{2}\cdot\frac{499}{999}

Agora vamos resolver E_2. Assim:

E_2=\dfrac{1}{\sqrt1+\sqrt2}+\dfrac{1}{\sqrt2+\sqrt3}+\ldots +\dfrac{1}{\sqrt{99}+\sqrt{100}}

Vamos racionalizar cada fração da soma. Assim:

E_2=\dfrac{1}{\sqrt1+\sqrt2}\cdot\dfrac{\sqrt1-\sqrt2}{\sqrt1-\sqrt2}+\dfrac{1}{\sqrt2+\sqrt3}\cdot\dfrac{\sqrt2-\sqrt3}{\sqrt2-\sqrt3}+\ldots +\dfrac{1}{\sqrt{99}+\sqrt{100}}\cdot\dfrac{\sqrt{99}-\sqrt{100}}{\sqrt{99}-\sqrt{100}}

E_2=\dfrac{\sqrt1-\sqrt2}{(\sqrt1)^2-(\sqrt2)^2}+\dfrac{\sqrt2-\sqrt3}{(\sqrt2)^2-(\sqrt3)^2}+\ldots +\dfrac{\sqrt{99}-\sqrt{100}}{(\sqrt{99})^2-(\sqrt{100})^2}

E_2=\dfrac{\sqrt1-\sqrt2}{1-2}+\dfrac{\sqrt2-\sqrt3}{2-3}+\ldots +\dfrac{\sqrt{99}-\sqrt{100}}{99-100}

E_2=\dfrac{\sqrt1-\sqrt2}{-1}+\dfrac{\sqrt2-\sqrt3}{-1}+\ldots +\dfrac{\sqrt{99}-\sqrt{100}}{-1}

E_2=-\sqrt1+\sqrt2-\sqrt2+\sqrt3+\ldots-\sqrt{99}+\sqrt{100}

E_2=\sqrt{100}-\sqrt1

E_2=10-1

E_2=9

Pronto! Agora basta juntar tudo. Assim:

E=E_1^{E_2}

E=\left(\frac{1001}{2}.\frac{499}{999}\right)^{9}