2022英语周报外研版第，答案

18.解:(1)设等比数列{an}的公比为q,由已知an 1=2S 2,可得=2S1-1 2(m≥2),两式相减可得an 1-an=25-251分即an 1-an=ga、整理得an 1=3an,可知q=3,…2分已知a 1=2S 2,令n=1,得a2=2a1 2,即a1q=2a1 2,解得a1=2,3分故等比数列{an)的通项公式为an=2·3(n∈N);4分由b1=2,(n 2)=m,(n∈N”)得一22,那么一号,一是一…一m2,一m以上m个式子相乘,可得一××号“2n 1n(n 1)b=n(n 1)(n≥2)5分又b1=2满足上式,所以(b。)的通项公式b=n(n 1)(n∈N).……6分(2)若在a。与an 1之间插入n个数,使这n 2个数组成一个公差为cn的等差数列,则a 1-an=(n 1)cn,……7分即为2,3-2·3=(m 1),整理得,1,所以b。=4,n3…,………………………………9分T=b1 b2C2 b23 … b-c4 b=4·1·3 4·2·32 4·3·32 … 4·(n-1)3-2 4·n·3-1=4(1·3° 2·32 3·32 … (n-1)32 n·3-1)3T=4[1·32 2·32 … (n-1)3 n3],10分两式相减得:2T=4(3 3 ·32所以=2(m,3 12x)=1 (2n-1)(m∈N).…分

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