Se aplica formula 1 /n·(n+k) =1/k ·(1/n -1 /n+k) ,unde n;k∈N\{0} a.i. 0 < k < n .
In cazul nostru avem
1 /2·5 +1 /5·8 +1 /8·11 +...+1 /(3n-1)·(3n+2) =1/3 ·(1/2 -1/5 +1/5 -1/8 +1/8 -1/11 +...+1 /3n-1 -1 /3n+2) =1/3 ·(1/2 -1 /3n+2) =1/3 ·3n+1 /2·(3n+2) =3n /18n+12 =n /6n+4 =n /2·(3n+2) ,ceea ce trebuia demonstrat .
