数列 (28 11:44:27) 已知数列{an},{bn}满足bn+1=bn+4an,bn=(2n-1)an求证:数列{an}为等差数列若a2=a1+b1-1,求∑(1/b)(要完整过程)
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/11 13:30:00
![数列 (28 11:44:27) 已知数列{an},{bn}满足bn+1=bn+4an,bn=(2n-1)an求证:数列{an}为等差数列若a2=a1+b1-1,求∑(1/b)(要完整过程)](/uploads/image/z/61776-0-6.jpg?t=%E6%95%B0%E5%88%97+%2828+11%3A44%3A27%29%26%23160%3B%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%2C%7Bbn%7D%E6%BB%A1%E8%B6%B3bn%2B1%3Dbn%2B4an%2Cbn%3D%282n-1%29an%E6%B1%82%E8%AF%81%EF%BC%9A%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E8%8B%A5a2%3Da1%2Bb1-1%2C%E6%B1%82%E2%88%91%281%2Fb%29%28%E8%A6%81%E5%AE%8C%E6%95%B4%E8%BF%87%E7%A8%8B%29)
数列 (28 11:44:27) 已知数列{an},{bn}满足bn+1=bn+4an,bn=(2n-1)an求证:数列{an}为等差数列若a2=a1+b1-1,求∑(1/b)(要完整过程)
数列 (28 11:44:27)
已知数列{an},{bn}满足bn+1=bn+4an,bn=(2n-1)an
求证:数列{an}为等差数列
若a2=a1+b1-1,求∑(1/b)
(要完整过程)
数列 (28 11:44:27) 已知数列{an},{bn}满足bn+1=bn+4an,bn=(2n-1)an求证:数列{an}为等差数列若a2=a1+b1-1,求∑(1/b)(要完整过程)
bn+1=bn+4an,bn=(2n-1)an,有b(n+1)=(2n+1)a(n+1),
所以,(2n+1)a(n+1)=(2n-1)an+4an=(2n+3)an,
进而有a(n+1)/an=(2n+3)/(2n+1),得
a2/a1=5/3,
a3/a2=7/5,
a4/a3=9/7,
……
an/a(n-1)=(2n+1)/(2n-1).
an/a1=(2n+1)/3,
an=(2n+1)a1/3,
a(n+1)-an=[(2n+3)-(2n+1)]a1/3
=2a1/3为常数,所以,数列{an}为等差数列.
a2=a1+b1-1,b1=a1,a2=5a1/3=5b1/3,可得b1=3=a1.
an=2n+1,bn=(2n-1)(2n+1).
“∑(1/b)”应该是“∑(1/bn)”吧?
∑(1/bn)=1/(1*3)+1/(3*5)+1/(5*7)+……+1/[(2n-1)(2n+1)]
=(1/2){[1/1-1/3]+[1/3-1/5]+[1/5-1/7}+……+[1/(2n-1)-1/(2n+1)]}
=(1/2)[1-1/(2n+1)]
=n/(2n+1).