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用数学归纳法证明,1-x/1!+x(x-1)/2!+...+(-1)^nx(x-1)...(x-n+1)/n!=(-1)
题目内容：
用数学归纳法证明,1-x/1!+x(x-1)/2!+...+(-1)^nx(x-1)...(x-n
+1)/n!=(-1)^n(x-1)(x-2)...(x-n)/n!
解惑(1)当n=1时,左边=(-1)x(x-1)(x-2)..x/1!,右边=(-1)(x-1)(x-2)..(x-1)/1!.左边x不等于右边x-1,怎样算才能左边=右边?
(2)当n=k+1时,为何(-1)^k(x-1)(x-2)..(x-k)/k!+(-1)^(k+1)x(x-1)(x-k)/(k+1)!不能直接相加?而要(-1)^(k+1)x(x-1)(x-k)/(k+1)!-(-1)^(k+1)(x-1)(x-2)..(x-k)/k!
用数学归纳法证明,1-x/1!+x(x-1)/2!+...+(-1)^nx(x-1)...(x-n
+1)/n!=(-1)^n(x-1)(x-2)...(x-n)/n!
解惑(1)当n=1时,左边=(-1)x(x-1)(x-2)..x/1!,右边=(-1)(x-1)(x-2)..(x-1)/1!.左边x不等于右边x-1,怎样算才能左边=右边?
(2)当n=k+1时,为何(-1)^k(x-1)(x-2)..(x-k)/k!+(-1)^(k+1)x(x-1)(x-k)/(k+1)!不能直接相加?而要(-1)^(k+1)x(x-1)(x-k)/(k+1)!-(-1)^(k+1)(x-1)(x-2)..(x-k)/k!
+1)/n!=(-1)^n(x-1)(x-2)...(x-n)/n!
解惑(1)当n=1时,左边=(-1)x(x-1)(x-2)..x/1!,右边=(-1)(x-1)(x-2)..(x-1)/1!.左边x不等于右边x-1,怎样算才能左边=右边?
(2)当n=k+1时,为何(-1)^k(x-1)(x-2)..(x-k)/k!+(-1)^(k+1)x(x-1)(x-k)/(k+1)!不能直接相加?而要(-1)^(k+1)x(x-1)(x-k)/(k+1)!-(-1)^(k+1)(x-1)(x-2)..(x-k)/k!
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