已知数列{a n}满足a1=2,a n+1=(3a n—2)/(2a n—1)求:
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已知数列{a n}满足a1=2,a n+1=(3a n—2)/(2a n—1)求:
(1)求出a2 a3的值,并求出数列{a n}的通项公式; (2)设bn=an/2n(2n+1),sn为数列{bn}的前n项和,证明sn
上面的a n+1是第n+1项 而后的(3a n—2)/(2a n—1)中的a n则是第n项。
(1)求出a2 a3的值,并求出数列{a n}的通项公式; (2)设bn=an/2n(2n+1),sn为数列{bn}的前n项和,证明sn
上面的a n+1是第n+1项 而后的(3a n—2)/(2a n—1)中的a n则是第n项。
![已知数列{a n}满足a1=2,a n+1=(3a n—2)/(2a n—1)求:](/uploads/image/z/17895587-59-7.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ba+n%7D%E6%BB%A1%E8%B6%B3a1%3D2%2Ca+n%2B1%3D%EF%BC%883a+n%E2%80%942%EF%BC%89%2F%EF%BC%882a+n%E2%80%941%EF%BC%89%E6%B1%82%EF%BC%9A)
1.
a2=(3a1-2)/(2a1-1)=(3×2-2)/(2×2-1)=4/3
a3=(3a2-2)/(2a2-1)=[3×(4/3)-2]/[2×(4/3)-1]=6/5
a1=2=2/1=(2×1)/(2×1-1)
a2=4/3=(2×2)/(2×2-1)
a3=6/5=(2×3)/(2×3-1)
猜想第n项an=(2n)/(2n-1)
证:
n=1时,a1=2=(2×1)/(2×1-1),满足表达式.
假设当n=k(k∈N+)时,表达式成立,即ak=(2k)/(2k-1),则当n=k+1时,
a(k+1)=(3ak-2)/(2ak-1)
=[3×(2k)/(2k-1) -2]/[2×(2k)/(2k-1) -1]
=[3×(2k)-2×(2k-1)]/[2×(2k)-(2k-1)]
=(2k+2)/(2k+1)
=[2(k+1)]/[2(k+1)-1],表达式同样成立.
k为任意正整数,因此数列{an}的通项公式为an=2n/(2n-1).
2.
bn=an/[2n(2n+1)]=2n/[2n(2n+1)(2n-1)]=1/[(2n+1)(2n-1)]=(1/2)[1/(2n-1) -1/(2n+1)]
Sn=b1+b2+...+bn
=(1/2)[1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1)]
=(1/2)[1-1/(2n+1)]
=n/(2n+1)
=1/(2 +1/n)
a2=(3a1-2)/(2a1-1)=(3×2-2)/(2×2-1)=4/3
a3=(3a2-2)/(2a2-1)=[3×(4/3)-2]/[2×(4/3)-1]=6/5
a1=2=2/1=(2×1)/(2×1-1)
a2=4/3=(2×2)/(2×2-1)
a3=6/5=(2×3)/(2×3-1)
猜想第n项an=(2n)/(2n-1)
证:
n=1时,a1=2=(2×1)/(2×1-1),满足表达式.
假设当n=k(k∈N+)时,表达式成立,即ak=(2k)/(2k-1),则当n=k+1时,
a(k+1)=(3ak-2)/(2ak-1)
=[3×(2k)/(2k-1) -2]/[2×(2k)/(2k-1) -1]
=[3×(2k)-2×(2k-1)]/[2×(2k)-(2k-1)]
=(2k+2)/(2k+1)
=[2(k+1)]/[2(k+1)-1],表达式同样成立.
k为任意正整数,因此数列{an}的通项公式为an=2n/(2n-1).
2.
bn=an/[2n(2n+1)]=2n/[2n(2n+1)(2n-1)]=1/[(2n+1)(2n-1)]=(1/2)[1/(2n-1) -1/(2n+1)]
Sn=b1+b2+...+bn
=(1/2)[1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1)]
=(1/2)[1-1/(2n+1)]
=n/(2n+1)
=1/(2 +1/n)
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