N=1 ⇒ S(1) = 1/6 ·1·2·3·2 = 2
n=2 ⇒ S(2) = 1/6 ·2·3·4·5 = 20
n=3 ⇒ S(3) = 1/6 ·3·4·5·8 = 80
Предположим что S(n) = 1/6n(n+1)(n+2)(3n-1) верно , доказать
S(n+1) = 1/6(n+1)(n+2)(n+3)·[3(n+1)-1] =
=1/6(n+1)(n+2)(n+3)·[(3n-1)+3] =
=1/6·(n+1)·(n+2)(n+3)·(3n-1) + 1/6·(n+1)(n+2)(n+3)·3 =
= 1/6·(n+1)(n+2)·n·(3n-1) +1/6·3(n+1)(n+2)(3n-1) +
+1/2(n+1)(n+2)(n+3) = S(n) + 1/2·(n+1)(n+2)·(3n-1) + +1/2(n+1)(n+2)(n+3) = S(n) + 1/2·(n+1)(n+2)·[(3n -1)+ +(n+3)]= S(n) +1/2·(n+1)(n+2)·(4n+2)=
= S(n) + (n+1)·(n+2)[2(n+1)-1]