原式=1991x1992x3983/6 =2632810796
看了图片 就明白为什么了
答案1991*1992*(2*1991+1)/ 6=15796864776
用数列求和规律!
Sn = 1² + 2² + 3² + ...... + n² = n(n+1)(2n+1)/6
归纳法证明:
n = 1, 1×(1+1)×(2×1+1)/6 = 6/6 = 1,求和公式正确
设 n = k 时,Sk = 1² + 2² + 3² + ...... + k² = k(k+1)(2k+1)/6 成立。
S(k+1) = k(k+1)(2k+1)/6+(k+1)²
= (k+1)[k(2k+1)/6+(k+1)]
= (k+1)[k(2k+1)+6k+6]/6
= (k+1)[2k²+7k+6]/6
= (k+1)[(k+2)(2k+3]/6
= (k+1)[(k+1)+1][2(k+1)+1]/6
得证。
答案1991*1992*(2*1991+1)/ 6=15796864776
Sn = 1² + 2² + 3² + ...... + n² = n(n+1)(2n+1)/6
归纳法证明:
n = 1, 1×(1+1)×(2×1+1)/6 = 6/6 = 1,求和公式正确
设 n = k 时,Sk = 1² + 2² + 3² + ...... + k² = k(k+1)(2k+1)/6 成立。
S(k+1) = k(k+1)(2k+1)/6+(k+1)²
= (k+1)[k(2k+1)/6+(k+1)]
= (k+1)[k(2k+1)+6k+6]/6
= (k+1)[2k²+7k+6]/6
= (k+1)[(k+2)(2k+3]/6
= (k+1)[(k+1)+1][2(k+1)+1]/6
你的问题可以展开为另一个题目:
1+2+3+...+1991
2+3+4+...+1991
3+4+5+...+1991
....
1989+1990+1991
1990+1991
1991
以上每一个求和结果再相加
=1991x1992x3983/6