Question

$\frac{1}{2} + \frac{1}{6 } + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} + ... + \frac{1}{n( +n + 1)} = \frac{2018}{2019}$
va rog sa ma ajutați ​

Answer 1

Răspuns:

Explicație pas cu pas:

1/2 + 1/6 + 1/12 + 1/20 + 1/30 + .....+ 1/n(n+1)  = 2018/2019

1/2 + 1/6 + 1/12 + 1/20 + 1/30 + .....+ 1/n(n+1) =

= 1/1·2 +1/2·3 + 1/3·4 + 1/4·5 + ....+ 1/n(n+1) =

= 1/1 -1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 -1/5 + .....+ 1/n  - 1/(n+1) =

= 1 - 1/(n+1)  = 2018/2019  =>

(n+1-1)/(n+1) = 2018/2019 =>

n/(n+1) = 2018/2019 => n = 2018