设等差数列{an}的前n项和为Sn,且S3=2S2+4,a5=36（1）求an,Sn（2）设bn=Sn-1（n∈N﹢）,Tn=1/b1+1/b2+1/b3+·····+1/bn,求Tn

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设等差数列{an}的前n项和为Sn,且S3=2S2+4,a5=36（1）求an,Sn（2）设bn=Sn-1（n∈N﹢）,Tn=1/b1+1/b2+1/b3+·····+1/bn,求Tn
设等差数列{an}的前n项和为Sn,且S3=2S2+4,a5=36
（1）求an,Sn
（2）设bn=Sn-1（n∈N﹢）,Tn=1/b1+1/b2+1/b3+·····+1/bn,求Tn

设等差数列{an}的前n项和为Sn,且S3=2S2+4,a5=36（1）求an,Sn（2）设bn=Sn-1（n∈N﹢）,Tn=1/b1+1/b2+1/b3+·····+1/bn,求Tn
1、由S3=2S2+4 a5=36得：
3a1+3*(3-1)*d/2=2*[2a1+2*(2-1)*d/2]+4
a1+4d=36
化简得：
d-a1=4
a1+4d=36
解之得：a1=4 d=8
则：an=4+(n-1)*8=8n-4
Sn=(4+8n-4)*n/2=4n^2
2、bn=Sn-1=4n^2-1=(2n-1)*(2n+1) 则：
Tn=1/(1*3)+1/(3*5)+.+1/[(2n-1)(2n+1)]
=1/2*[1-1/3+1/3-1/5+.+1/(2n-1)-1/(2n+1)]
=1/2*[1-1/(2n+1)]
=n/(2n+1)

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