3 S = 1 ⋅ 2 ⋅ 3 + 2 ⋅ 3 ⋅ ( 4 − 1 ) + 3 ⋅ 4 ⋅ ( 5 − 2 ) + … + 99 ⋅ 100 ⋅ ( 101 − 98 ) . 3S=1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot (4-1) + 3 \cdot 4 \cdot (5-2) + \ldots + 99 \cdot 100 \cdot (101-98). 3 S = 1 ⋅ 2 ⋅ 3 + 2 ⋅ 3 ⋅ ( 4 − 1 ) + 3 ⋅ 4 ⋅ ( 5 − 2 ) + … + 99 ⋅ 100 ⋅ ( 101 − 98 ) . 3 S = 1 ⋅ 2 ⋅ 3 + ( 2 ⋅ 3 ⋅ 4 − 1 ⋅ 2 ⋅ 3 ) + ( 3 ⋅ 4 ⋅ 5 − 2 ⋅ 3 ⋅ 4 ) + … + ( 99 ⋅ 100 ⋅ 101 − 98 ⋅ 99 ⋅ 100 ) . 3S=\cancel{1 \cdot 2 \cdot 3} + (\cancel{2 \cdot 3 \cdot 4} - \cancel{1 \cdot 2 \cdot 3}) + (3 \cdot 4 \cdot 5 -\cancel{2 \cdot 3 \cdot 4}) + \ldots + (99 \cdot 100 \cdot 101-\cancel{98 \cdot 99 \cdot 100}). 3 S = 1 ⋅ 2 ⋅ 3 + ( 2 ⋅ 3 ⋅ 4 − 1 ⋅ 2 ⋅ 3 ) + ( 3 ⋅ 4 ⋅ 5 − 2 ⋅ 3 ⋅ 4 ) + … + ( 99 ⋅ 100 ⋅ 101 − 98 ⋅ 99 ⋅ 100 ) . S = 99 ⋅ 100 ⋅ 101 : 3. S=99 \cdot 100 \cdot 101 : 3. S = 99 ⋅ 100 ⋅ 101 : 3.
Metoda 2 (opțional): Ne folosim de egalitatea n ( n + 1 ) = n 2 + n n(n+1) = n^2+n n ( n + 1 ) = n 2 + n 1 2 + 2 2 + … + n 2 = n ( n + 1 ) ( 2 n + 1 ) : 6 . \boxed{1^2+2^2+ \ldots + n^2 = n(n+1)(2n+1):6}. 1 2 + 2 2 + … + n 2 = n ( n + 1 ) ( 2 n + 1 ) : 6 . S = ( 1 2 + 1 ) + ( 2 2 + 2 ) + ( 3 2 + 3 ) + … + ( 99 2 + 99 ) = S=(1^2+1) + (2^2+2) + (3^2+3) + \ldots + (99^2+99)= S = ( 1 2 + 1 ) + ( 2 2 + 2 ) + ( 3 2 + 3 ) + … + ( 9 9 2 + 99 ) = ( 1 2 + 2 2 + … + 99 2 ) + ( 1 + 2 + … + 99 ) = (1^2+2^2+ \ldots + 99^2) + (1+2+\ldots + 99)= ( 1 2 + 2 2 + … + 9 9 2 ) + ( 1 + 2 + … + 99 ) = = 99 ⋅ 100 ⋅ 199 : 6 + 99 ⋅ 100 : 2 = =99 \cdot 100 \cdot 199:6 + 99 \cdot 100 :2= = 99 ⋅ 100 ⋅ 199 : 6 + 99 ⋅ 100 : 2 = = 99 ⋅ 100 ⋅ 101 : 3. =99 \cdot 100 \cdot 101 : 3. = 99 ⋅ 100 ⋅ 101 : 3.