神马作文网数列 (28 11:44:27)
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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)
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)
n+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).
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).