29.5)用数学归纳法证明:1- (1/2)+ (1/3) -(1/4)+.+(1/(2n-1))-(1/(2n))=(1/(n+1))+(1/((n+2))+...+(1/2n)(n属于N*)

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/29 13:28:06

29.5)用数学归纳法证明:1- (1/2)+ (1/3) -(1/4)+.+(1/(2n-1))-(1/(2n))=(1/(n+1))+(1/((n+2))+...+(1/2n)(n属于N*)
29.5)
用数学归纳法证明:
1- (1/2)+ (1/3) -(1/4)+.+(1/(2n-1))-(1/(2n))=(1/(n+1))+(1/((n+2))+...+(1/2n)
(n属于N*)

29.5)用数学归纳法证明:1- (1/2)+ (1/3) -(1/4)+.+(1/(2n-1))-(1/(2n))=(1/(n+1))+(1/((n+2))+...+(1/2n)(n属于N*)
当n=1时 显然成立,
考虑n大于1,假设k使上式成立,即有[1+1/3+.1/(2k-1)]-[1/2+1/4+
.+1/2k]=[1/(k+1)+(1/(k+2)+...+(1/2k)]
下面考虑k+1,此时左边有[1+1/3+.1/(2k-1)+1/(2k+1)]-[1/2+1/4+.+1/2k+1/(2k+2)]=[1/(k+1)+...+(1/2k)]+
1/(2k+1)-1/(2k+2),将首末两项结合可得左边为
1/(k+2)+1/(k+3)+...+1/(2k+1)+1/(2k+2)
至此 命题得证

简单,只要证1/(2(n+1)*(2n+1)) + 1/(n+1)=1/(2n+1)+1/(2n+2)
慢慢想想吧,动动脑筋,别老上来求助,大一师兄留言

注意当n+1时有左边1- (1/2)+ (1/3) -(1/4)+....+(1/(2n-1))-(1/(2n)+1/[2n+1]-1/[2n+2]只需证明它等于1/[n+2]+1/[n+2]+...+...1/[2n]+1/[2n+1]+1/[2n+2]可以归结为证明如二楼所说的1/(2(n+1)*(2n+1)) + 1/(n+1)=1/(2n+1)+1/(2n+2)
,显然的容易证明,...

全部展开

注意当n+1时有左边1- (1/2)+ (1/3) -(1/4)+....+(1/(2n-1))-(1/(2n)+1/[2n+1]-1/[2n+2]只需证明它等于1/[n+2]+1/[n+2]+...+...1/[2n]+1/[2n+1]+1/[2n+2]可以归结为证明如二楼所说的1/(2(n+1)*(2n+1)) + 1/(n+1)=1/(2n+1)+1/(2n+2)
,显然的容易证明,可以先把 1/(n+1)移到右边,和1/(2n+2)配对,因为有公约式,

收起

xy需要细节加我