等差数列an中,a3=9,a5=17,记数列1/an的前n和是Sn,S2n+1-Sn小于等于1/10m,求整数m最小值_数学解答

等差数列an中,a3=9,a5=17,记数列1/an的前n和是Sn,S2n+1-Sn小于等于1/10m,求整数m最小值

人气：448 ℃　时间：2020-07-27 20:01:48

解答

因为a3=9,a5=17所以解得a1=1,d=4所以an=4n-3故Sn=1+1/5+1/9+…+1/(4n-3)令Hn=S(2n+1)-Sn=1/(4n+1)+…+1/(8n+1)H(n+1)-Hn=1/[4(n+1)+1]+…+1/[8(n+1)+1]-[1/(4n+1)+…+1/(8n+1)]=1/(8n+5)+1/(8n+9)-1/(4n+1)...