IME / ITA ⇒ (Colégio Naval - 1994) Séries Tópico resolvido

[alinebotelho]

Sabendo-se que a seguinte identidade [tex3]\Large\frac{a\,\cdot\,x+b\,\cdot\,y}{x\,\cdot\,y}\large=\Large\frac{a}{y}\large + \Large\frac{b}{x}\large[/tex3] é verdadeira para quaisquer números reais [tex3]a,\,b,[/tex3] [tex3]x\neq 0[/tex3] e [tex3]y\neq 0 ,[/tex3] o valor de [tex3]\frac{13}{2\,\cdot\,4}+\frac{13}{4\,\cdot\,6}+\frac{13}{6\,\cdot\,8}+\ldots+\frac{13}{50\,\cdot\,52}[/tex3] é igual a:


a) [tex3]\frac{25}{16}[/tex3]
b) [tex3]\frac{25}{12}[/tex3]
c) [tex3]\frac{25}{8}[/tex3]
d) [tex3]\frac{25}{4}[/tex3]
e) [tex3]\frac{25}{2}[/tex3]

[agp16]

• [tex3]\frac{13}{2\,\cdot\,4}+\frac{13}{4\,\cdot\,6}+\frac{13}{6\,\cdot\,8}+\ldots+\frac{13}{50\,\cdot\,52}=\frac{13}{4}\cdot\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\ldots+\frac{1}{650}\right)[/tex3]

Como

• [tex3]\begin{array}{rl} \frac{1}{2}&=1-\frac{1}{2}\\
  \frac{1}{6}&=\frac{1}{2}-\frac{1}{3} \\
  \frac{1}{12}&=\frac{1}{3}-\frac{1}{4} \\
  &\vdots \\
  \frac{1}{650}&=\frac{1}{25}-\frac{1}{26} \\
  \end{array}[/tex3]

temos

• [tex3]\begin{array}{rl} \frac{13}{4}\cdot\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\ldots+\frac{1}{650}\right)&=\frac{13}{4}\cdot\left(1-\cancel{\frac{1}{2}}+\cancel{\frac{1}{2}}-\cancel{\frac{1}{3}}+\cancel{\frac{1}{3}}-\cancel{\frac{1}{4}}+\ldots+\cancel{\frac{1}{25}}-\frac{1}{26}\right)\\
  &=\frac{13}{4}\cdot\left(1- \frac{1}{26}\right)\\
  &=\frac{13}{4}\cdot\frac{25}{26} \\
  &=\frac{25}{8}.\end{array}[/tex3]