已知数列Sn=2An+(-1)^n n大于一 试证明对于任意m大于4有 1/A4 +1/A5 +1/A6 +.+1/Am 小于7/8我算出An=[2^(n-1)-2(-1)^n]/3
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/04 19:57:56
![已知数列Sn=2An+(-1)^n n大于一 试证明对于任意m大于4有 1/A4 +1/A5 +1/A6 +.+1/Am 小于7/8我算出An=[2^(n-1)-2(-1)^n]/3](/uploads/image/z/5457533-5-3.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97Sn%3D2An%2B%EF%BC%88-1%29%5En+n%E5%A4%A7%E4%BA%8E%E4%B8%80+%E8%AF%95%E8%AF%81%E6%98%8E%E5%AF%B9%E4%BA%8E%E4%BB%BB%E6%84%8Fm%E5%A4%A7%E4%BA%8E4%E6%9C%89+1%2FA4+%2B1%2FA5+%2B1%2FA6+%2B.%2B1%2FAm+%E5%B0%8F%E4%BA%8E7%2F8%E6%88%91%E7%AE%97%E5%87%BAAn%3D%5B2%5E%28n-1%29-2%28-1%29%5En%5D%2F3)
已知数列Sn=2An+(-1)^n n大于一 试证明对于任意m大于4有 1/A4 +1/A5 +1/A6 +.+1/Am 小于7/8我算出An=[2^(n-1)-2(-1)^n]/3
已知数列Sn=2An+(-1)^n n大于一 试证明对于任意m大于4有 1/A4 +1/A5 +1/A6 +.+1/Am 小于7/8
我算出An=[2^(n-1)-2(-1)^n]/3
已知数列Sn=2An+(-1)^n n大于一 试证明对于任意m大于4有 1/A4 +1/A5 +1/A6 +.+1/Am 小于7/8我算出An=[2^(n-1)-2(-1)^n]/3
An=[2^(n-1)-2(-1)^n]/3 =(2/3)[2^(n-2)+(-1)^(n-1)].
A4=2.
若m为偶数,则
1/A4 +1/A5 +1/A6 +...+1/Am
=1/A4 +(1/A5+1/A6)+...+[1/A(m-1)+1/Am]
其中1/A(m-1)+1/Am
=(3/2){1/[2^(m-3)+1]+1/[2^(m-2)-1]}
=(3/2){[2^(m-3)+1+2^(m-2)-1]/[2^(m-3)+1][2^(m-2)-1]}
=(3/2){[2^(m-3)+2^(m-2)]/[2^(2m-5)-2^(m-3)+2^(m-2)-1]
<(3/2){[2^(m-3)+2^(m-2)]/[2^(2m-5)][∵-2^(m-3)+2^(m-2)-1>0]
=(3/2)[1/2^(m-3)+1/2^(m-2)]
∴1/A4 +1/A5 +1/A6 +...+1/Am
=1/A4 +(1/A5+1/A6)+...+(1/A(m-1)+1/Am)
<(1/2)+(3/2)[1/2^3+1/2^4+1/2^5+...+1/2^(m-2)]
=(1/2)+(3/2)(1/4)[1-1/2^(m-4)]
<(1/2)+(3/8)
=7/8.
当m是奇数时,m+1是偶数,所以
1/A4 +1/A5 +1/A6 +...+1/Am
<1/A4 +1/A5 +1/A6 +...+1/Am+1/A(m+1)
<7/8.
综上,命题得证.