等差数列{a n }中,首项a 1 =1,公差d≠0,前n项和为S n ,已知数列 , , ,…, ,…成等比数列,

2025-04-04 21:51:51
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解:(Ⅰ)由a 2 2 =a 1 a 5 ,得(1+d)2=1(1+4d),
解得d=2,
∴a n =2n﹣1,
=2k n ﹣1,在等比数列中,公比q= =3,
=3 n﹣1 ,∴2k n ﹣1=3 n﹣1 ,解得k n =
(Ⅱ)b n = = ,则
+ +…+
T n = +…+ +
两式相减得: T n =1+ +…+ =2﹣
∴T n =3﹣
∵T n+1 ﹣T n = >0,
∴T n 单调递增,
∴1≤Tn<3.
在n∈N*时单调递增.
且S 1 =1,4T 1 =4;S 2 =4,4T 2 =8;S 3 =9,4T 3 = ;S 4 =16>12,4T 4 <12;….
n>3时,S n >4T n 恒成立,则所求最小正整数M的值为3.