Question

等差數(shù)列{an}中,2a1+3a2=11,2a3=a2+a6-4,其前n項(xiàng)和為Sn.

(1)求數(shù)列{an}的通項(xiàng)公式.

(2)設(shè)數(shù)列{bn}滿足bn=,其前n項(xiàng)和為Tn,求證:Tn<(n∈N*).

 

Answer 1

(1) an=2n-1 (2)見解析

【解析】(1)2a1+3a2=2a1+3(a1+d)=5a1+3d=11,

2a3=a2+a6-4,

即2(a1+2d)=a1+d+a1+5d-4,得d=2,

則a1=1,故an=2n-1.

(2)由(1)得Sn=n2,∴bn==

===(-),

Tn=(-+-+-+…+-+-)

=(+--)<(n∈N*).

【方法技巧】裂項(xiàng)相消法的應(yīng)用技巧

裂項(xiàng)相消法的基本思想是把數(shù)列的通項(xiàng)an分拆成an=bn+1-bn或者an=bn-bn+1或者an=bn+2-bn等,從而達(dá)到在求和時(shí)逐項(xiàng)相消的目的,在解題中要善于根據(jù)這個(gè)基本思想變換數(shù)列an的通項(xiàng)公式,使之符合裂項(xiàng)相消的條件.在裂項(xiàng)時(shí)一定要注意把數(shù)列的通項(xiàng)分拆成的兩項(xiàng)一定是某個(gè)數(shù)列中的相鄰的兩項(xiàng)或者是等距離間隔的兩項(xiàng),只有這樣才能實(shí)現(xiàn)逐項(xiàng)相消后剩下幾項(xiàng),達(dá)到求和的目的.