已知数列{an}（n∈N*）满足a1=1，an+1=3an+2．（Ⅰ）证明{an+1}是等比数列，并求{an}的通项公式；（Ⅱ）若数列{bn}满足bn=log3an+12，记Tn=1b2b4+1b3b5+1b4b6+…+1bnbn+2，求Tn．

Question

已知数列{a_n}（n∈N^*）满足a₁=1，a_n+1=3a_n+2．
（Ⅰ）证明{a_n+1}是等比数列，并求{a_n}的通项公式；
（Ⅱ）若数列{b_n}满足b_n=log₃(an+1)/(2)，记T_n=(1)/(b2b4)+(1)/(b3b5)+(1)/(b4b6)+…+(1)/(bnbn+2)，求T_n．

Answer 1

（Ⅰ）证明：∵a_n+1=3a_n+2，∴a_n+1+1=3（a_n+1）．
∴{a_n+1}是等比数列，首项为2，公比为3．
a_n+1=2×3^n-1，解得a_n=2×3^n-1-1．
（Ⅱ） 由（Ⅰ）知b_n=log₃(an+1)/(2)=n-1，
∴(1)/(bnbn+2)=(1)/((n-1)(n+1))=(1)/(2)((1)/(n-1)-(1)/(n+1))．
∴T_n=(1)/(b2b4)+(1)/(b3b5)+(1)/(b4b6)+…+(1)/(bnbn+2)=(1)/(2)[(1-(1)/(3))+((1)/(2)-(1)/(4))+((1)/(3)-(1)/(5))+…+((1)/(n-2)-(1)/(n))+((1)/(n-1)-(1)/(n+1))]
=(1)/(2)(1+(1)/(2)-(1)/(n)-(1)/(n+1))
=(3)/(4)-(2n+1)/(2n2+2n)．