已知等差数列{a n }的前n项和为S n ,且a 3 =5,S 6 =36.(Ⅰ)求数列{a n }的通项a n ;(Ⅱ)设 b

2025-05-13 17:36:18
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回答1:

(Ⅰ)由
a 3 = a 1 +2d=5
s 6 =6 a 1 +15d=36
解得
a 1 =1
d=2

∴a n =1+(n-1)d
(Ⅱ)b n =2 n
∴{b n }是以2为首项,2为公比的等比数列
∴T n =b 1 +b 2 ++b n =2+2 2 +2 3 ++2 n =2 n+1 -2