Question

sa se demonstreze identitatea
$1 + (1 + 2) + (1 + 2 + 3) + ... + (1 + 2 + 3 + ... + n) = \frac{n \times (n + 1) \times (n + 2)}{6}$
oricare ar fi n=nr natural

Answer 1

vom folosi sumele cunoscute:

1+2+3+...+n=n(n+1)/2

1²+2²+3³+....+n²=n(n+1)(2n+1)/6

Suma noastra se poate scrie:

1*2/2+2*3/2+3*4/2+....+n(n+1)/2=1/2[1(1+1)+2(1+2)+3(1+3)+....n(1+n)]=1/2[1+1²+2+2²+3+3²+...n+n²]=1/2[n(n+1)/2+n(n+1)(2n+1)/2]=1/4*n(n+1)[1+2n+1]=n(n+1)²/2