Question

已知数列{an}的前n项和为sn=n²+2n,数列{bn}是正项等比数列

已知数列{an}的前n项和为sn=n²+2n,数列{bn}是正项等比数列，且满足a1=2b1，b3（a3-a1）=b1，n∈正实数
（1）求数列{an}和{bn}的通项公式
（2）记cn=anbn，求数列{cn}的前n项的和

Answer 1

bn=1/[(2n+1)²-1]
=1/(4n²+4n)
=1/4[1/n-1/(n+1)]
∴Tn
=(b1+b2+b3+……+bn)
=1/4[(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/n-1/n+1)]
=1/4[1-1/(n+1)]
=1/4[(n+1-1)/(n+1)
=n/4(n+1)