(1)因为点(an+1,S2n-1)在函数f(x)的图象上,所以 =S2n-1. 令n=1,n=2,得 即 解得a1=1,d=2(d=-1舍去),则an=2n-1. 由(bn-bn+1)·g(bn)=f(bn), 得4(bn-bn+1)(bn-1)=(bn-1)2. 由题意bn≠1,所以4(bn-bn+1)=bn-1, 即3(bn-1)=4(bn+1-1),所以![](http://img.shitiku.com.cn/uploads/allimg/20191009/20191009064417-76380.png) 所以数列{bn-1}是以1为首项,公比为 的等比数列. (2)由(1),得bn-1= n-1.cn= . 令Tn=c1+c2+c3+…+cn, 则Tn= + + +…+ + ,①
Tn= + + +…+ + ,② ①-②得, Tn= + + + +…+ - =1+ · - =2- - =2- .所以Tn=3-.![](http://img.shitiku.com.cn/uploads/allimg/20191009/20191009064421-31923.png) 所以c1+c2+c3+…+cn=3- <3. |