数列an的前n项和为Sn,且Sn=2n的平方+4n,设数列bn的前n项和为Tn,bn=2\【an(2n-1)】,求Tn.
2个回答
s1=a1=2*1^2+4*1=6
sn=2n^2+4n
s(n-1)=2(n-1)^2+4(n-1)
=2n^2-4n+2+4n-4
=2n^2-2
an=sn-s(n-1)
=2n^2+4n-(2n^2-2)
=4n+2
bn=2/[an*(2n-1)]
=2/[(4n+2)*(2n-1)]
=1/[(2n-1)(2n+1)]
=1/2*[1/(2n-1)-1/(2n+1)]
b1=1/2*(1/1-1/3)
b2=1/2*(1/3-1/5)
.
bn=1/2*[1/(2n-1)-1/(2n+1)]
Tn=b1+b2+b3+.+bn
=1/2*[1-1/3]+1/2*[1/3-1/5]+1/2*[1/5-1/7]+.+1/2*[1/(2n-1)-1/(2n+1)]
=1/2*[1-1/3+1/3-1/5+1/5-1/7+.1/(2n-1)-1/(2n+1)]
=1/2*[1-1/(2n+1)]
=1/2*2n/(2n+1)]
=n/(2n+1)
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