俊裔随威
幼苗
共回答了14个问题采纳率:92.9% 举报
(1)设a n =a 1 +(n-1)d,S n =
n( a 1 + a n )
2 ,
所以 a 3 =a 1 +2d=5 ①,
S 15 =
15( a 1 + a 15 )
2 =15(a 1 +7d)=225
a 1 +7d=15 ②
①②联立解得d=2,a 1 =1,
∴数列{a n }的通项公式为a n =2n-1
设b n =b 1 •q (n-1) ,
所以 b 3 =a 2 +a 3 =8,
b 2 =
b 3
q ,b 5 =b 3 •q 2
∴b 2 •b 5 =b 3 2 •q=64•q=128
∴q=2
∴数列{b n }的通项公式为b n =b 3 •q n-3 =2 n (n=1,2,3,…).
(2)∵c n =(2n-1)•2 n
∵Tn=2+3•2 2 +5•2 3 +…+(2n-1)•2 n
2Tn=2 2 +3•2 3 +5•2 4 +…+(2n-3)•2 n +(2n-1)•2 n+1
作差:-Tn=2+2 3 +2 4 +2 5 +…+2 n+1 -(2n-1)•2 n+1
=2+23(1-2n-1)1-2-(2n-1)•2n+1
=2+
2 3 (1- 2 n-1 )
1-2 -(2n-1)•2 n+1
=2+2 n+2 -8-2 n+2 n+2 n+1 =-6-2 n+1 •(2n-3)
∴Tn=(2n-3)•2 n+1 +6(n=1,2,3,…).
1年前
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