若数列{bn}满足b(n+1)-bn=2n+3
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若数列{bn}满足b(n+1)-bn=2n+3,n为正整数,b1=3,求数列{1/bn}的前n项和Tn
b(n+1)-bn=2n+3,n为正整数,b1=3
可得
bn-b(n-1)=2(n-1)+3
b(n-1)-b(n-2)=2(n-2)+3
----
b2-b1=2+3
相加得
bn-b1=2+4+6+----+2(n-1)+3(n-1)=n^2+2n-3
bn=n^2+2n
1/bn=1/n(n+2)=(1/2)*[1/n-1/(n+2)]
数列{1/bn}的前n项和
Tn=(1/2)*[1-1/3+1/2-1/4+1/3-1/5+----+1/(n-1)-1/(n+1)+1/n-1/(n+2)]
=(1/2)*[1+1/2-1/(n+1)-1/(n+2)]
=3/4-(2n+3)/[2(n+1)(n+2)]
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