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(1/2)数列1*n,2(n-1),3(n-2),…,n*1的和为（）A.1/6n(n+1)(n+2)B.1/6n(n+1)(2
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(1/2)数列1*n,2(n-1),3(n-2),…,n*1的和为（）A.1/6n(n+1)(n+2) B.1/6n(n+1)(2
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答案和解析
n * 1 = n(n+1) - n²
1*n + 2(n-1) + 3(n-2) + ... + n*1
= (1+2+3+...+n)(n+1) - (1² + 2² + 3² + ... + n²)
= n(n+1)(n+1)/2 - n(n+1)(2n+1)/6
= n(n+1)(n+2)/6
1*n + 2(n-1) + 3(n-2) + ... + n*1
= (1+2+3+...+n)(n+1) - (1² + 2² + 3² + ... + n²)
= n(n+1)(n+1)/2 - n(n+1)(2n+1)/6
= n(n+1)(n+2)/6
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