Question

已知数列{an}的前n项和为sn,满足a(n+1)=sn+2n 且a1=0

（1）若bn=an+2求证{bn}是等比数列。（2）若C若n=1
／log2(bn+1)log2(bn) 求数列{Cn}的前你n项和Tn

Answer 1

证明：（1）an+1 = Sn + 2n①，当n ≥ 2，n∈N* ，an = Sn-1 + 2(n – 1)②，把① - ②，可得an+1 – an = Sn – Sn-1 + 2 = an + 2 => an+1 = 2an + 2，而bn = an + 2 => an = bn – 2 => an+1 = bn+1 – 2 => bn+1 – 2 = 2(bn – 2) + 2 = 2bn – 2 => bn+1 = 2bn => bn+1 /bn = 2，所以数列{b_n}构成一个以b1 = a1 + 2 = 2为首项，2为公比的等比数列，得证；
解：（2）由（1）可知bn = 2*2n-1 = 2n ，n∈N* ，Cn = 1/[log2(bn+1)log2(bn)] = 1/[log2(2n+1)log2(2n)] = 1/[n(n+1)] = 1/n – 1/(n + 1)，所以数列{C_n}的前n项的和Tn = C1 + C2 + …… + Cn = (1/1 – 1/2) + (1/2 – 1/3) + …… + [1/n – 1/(n + 1)] = 1 – 1/(n + 1) = n/(n + 1) ，即Tn = n/(n + 1)，n∈N* 。