Question

Помогите вычислить, пожалуйста!
$\frac{1}{15} + \frac{1}{35} + \frac{1}{63} + \frac{1}{99} + \frac{1}{143} + \frac{1}{195} =$

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Answer 1

1/n(n+2) = 1/2*(n+2 - n)/n(n+2) = 1/2*(1/n - 1/(n+2))

1/15 + 1/35 + 1/63 + 1/99 + 1/143 + 1/195 = 1/3*5 + 1/5*7 + 1/7*9 + 1/9*11 + 1/11*13 + 1/13*15 = 1.2*(1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13 + 13 - 1/15) = 1/2*(1/3 - 1/15) = 1/2*4/15 = 2/15

Answer 2

Телескопические суммы!

$S=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{195}\\S=2\left (\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...-\frac{1}{15} \right )\\S=2\cdot \left ( \frac{1}{3}-\frac{1}{15} \right )=\frac{2}{14}$