數學知識交流---求值(3)

1. ∑(n k=1) k^2 = n(n+1)(2n+1)/6

17^2 + 18^2 + ... + 98^2
= (1^2 + 2^2 + ... + 98^2) - (1^2 + 2^2 + ... + 16^2)
= 98(98+1)(2*98+1)/6 - 16(16+1)(2*16+1)/6
= 318549 - 1496
= 317053

2. ∑(n k=1) k^3
= [∑(n k=1) k]^2
= [n(n+1) / 2]^2
= n^2 (n+1)^2 / 4

1^3 + 2^3 + ... + 27^3
= 27^2 (27+1)^2 / 4
= 142884

3. 3^2 + 6^2 + 9^2 + ... + (3n)^2
= (1*3)^2 + (2*3)^2 + (3*3)^2 + ... + (3*n)^2
= 3^2 (1^2 + 2^2 + 3^2 + ... + n^2)
= 9(1^2 + 2^2 + 3^2 + ... + n^2)
= 9∑(n k=1) k^2
= 9n(n+1)(2n+1)/6
= 3n(n+1)(2n+1)/2

18^2 + 21^2 + 24^2 + ... + 195^2
= (3^2 + 6^2 + 9^2 + ... + 195^2) - (3^2 + 6^2 + 9^2 + 12^2 + 15^2)
= 3(65)(65+1)(2*65+1) / 2 - 3(5)(5+1)(2*5+1) / 2
= 842985 - 495
= 842490

(1) 求17²+ 18²+ 19²+ ... + 97²+ 98²的值。
解：17²+ 18²+ 19²+ ... + 97²+ 98²
= (1²+ 2²+ 3²+ ... + 98²) - (1²+ 2²+ 3²+ ... + 16²)
= [ 98×(98+1)×(98×2+1)/6 ] - [ 16×(16+1)×(16×2+1)/6 ]
= 318549 - 1496
= 317053

2011-07-03 19:25:50 補充：
(2) 求1³+ 2³+ 3³+ ... + 27³的值。
解：1³+ 2³+ 3³+ ... + 27³
= {[(1+27)×27]/2}²
= (27×14)²
= 142884

2011-07-03 19:25:53 補充：
(3) 求18² + 21² + 24² + ... + 192²+ 195²的值。
解：18² + 21² + 24² + ... + 192²+ 195²
= (3²× 6²) + (3²× 7²) + ... + (3²× 65²)
= 3²× (6²+ 7²+ ...... + 65²)
= 3²× [(1²+ 2²+ 3²+ ... + 65²) - (1²+ 2²+ 3²+ 4²+ 5²)]
= 3²× [65×(65+1)×(65×2+1)/6 - 5×(5+1)×(5×2+1)/6]
= 3²× (93665 - 55)
= 842490