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,
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/02 01:27:05
![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,](/uploads/image/z/5456923-43-3.jpg?t=%E7%94%A8%E6%95%B0%E5%AD%A6%E5%BD%92%E7%BA%B3%E6%B3%95%E8%AF%81%E6%98%8E%2C1-x%2F1%21%2Bx%28x-1%29%2F2%21%2B...%2B%28-1%29%5Enx%28x-1%29...%28x-n%2B1%29%2Fn%21%3D%28-1%29%5En%28x-1%29%28x-2%29...%28x-n%29%2Fn%21%E8%A7%A3%E6%83%91%281%29%E5%BD%93n%3D1%E6%97%B6%2C%E5%B7%A6%E8%BE%B9%3D%28-1%29x%28x-1%29%28x-2%29..x%2F1%21%2C%E5%8F%B3%E8%BE%B9%3D%28-1%29%28x-1%29%28x-2%29..%28x-1%29%2F1%21.%E5%B7%A6%E8%BE%B9x%E4%B8%8D%E7%AD%89%E4%BA%8E%E5%8F%B3%E8%BE%B9x-1%2C%E6%80%8E%E6%A0%B7%E7%AE%97%E6%89%8D%E8%83%BD%E5%B7%A6%E8%BE%B9%3D%E5%8F%B3%E8%BE%B9%3F%282%29%E5%BD%93n%3Dk%2B1%E6%97%B6%2C)
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,
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,
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