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1A+(1A+2)A+(2A+3)A+(3A+4)A+...+(AA+A+1)A=(1)A+(2)A+(3)A+(4)A+...+(AA)A How do you prove it ? Did A can be everything? If not , a = ? PleasePleasePlease!!! 更新: If a=1 1+1+2=1 4=1 ??? A can't be everything 更新 2: ha ha , you're right!

最佳解答:

1A+(1A+2)A+(2A+3)A+(3A+4)A+...+(AA+A+1)A=(1)A+(2)A+(3)A+(4)A+...+(AA)A 先看左面,先考慮以下數列的通項: T(n) = (nA + n + 1)A T(0) = 1A T(1) = (1A + 2)A T(2) = (2A + 3)A T(A) = (AA + A + 1)A 因此 1A+(1A+2)A+(2A+3)A+(3A+4)A+...+(AA+A+1)A = T(0) + T(1) + T(2) + ... + T(A) = ∑(n = 0 to A) T(n) = ∑(n = 0 to A) (nA + n + 1)A = A^2[∑(n = 0 to A) n] + A[∑(n = 0 to A) n] + A[∑(n = 0 to A) 1] = (A^2)(A)(A + 1)/2 + (A)(A)(A + 1)/2 + A(A + 1) = A(A + 1)(A^2 + A + 2)/2 看右面 (1)A+(2)A+(3)A+(4)A+...+(AA)A = ∑(n = 1 to AA) nA = A[∑(n = 1 to AA) n] = A(AA)(AA + 1)/2 = A^3(A^2 + 1)/2 要左右相等, A(A + 1)(A^2 + A + 2)/2 = A^3(A^2 + 1)/2 A(A + 1)(A^2 + A + 2) = A^3(A^2 + 1) (A^2 + A)(A^2 + A + 2) = A^5 + A^3 A^4 + A^3 + 2A^2 + A^3 + A^2 + 2A = A^5 + A^3 A^5 - A^4 - A^3 - 3A^2 - 2A = 0 A(A^4 - A^3 - A^2 - 3A - 2) = 0 括號中的正整數實根只有１或２兩個可能。但1 - 1 - 1 - 3 - 2不等於０，16 - 8 - 4 - 6 - 2也不等於０。右面的數列是由１開始的，所以0不是答案,這道題沒有解。

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