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,

Question

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!

Answer 1

前面n=1时式子成立不写了
假设n=k成立则1/x!+.(-1)^k x(x-1)(x-k+1)/k!=(-1)^k (x-1)(x-2)...(x-k)/k!成立
则n=k+1时有1/x!+.(-1)^k x(x-1)(x-k+1)/k!+(-1)^(k+1) x(x-1)(x-k)/(k+1)!=(-1)^k (x-1)(x-2)...(x-k)/k!+(-1)^(k+1) x(x-1)(x-k)/(k+1)!=(-1)^(k+1) x(x-1)(x-k)/k!(k+1) - (-1)^(k+1) (x-1)(x-2)...(x-k)/k!=(-1)^(k+1) (x-1)(x-2)...(x-k)/k!*[x/(k+1)-1]=(-1)^(k+1) (x-1)(x-2)[x-(k+1)]/k!(k+1)=(-1)^(k+1) (x-1)(x-2)[x-(k+1)]/(k+1)!即当n=k+1时也成立；故式子得证