庄菲
花朵
共回答了16个问题采纳率:81.3% 举报
(1)当n=1时,a 1 =S 1 =2a 1 -2,解得a 1 =2.
当n≥2时,a n =S n -S n-1 =2a n -2-(2a n-1 -2),
化为a n =2a n-1 ,
∴数列{a n }是以2为首项,2为公比的等比数列.
∴ a n = 2 n .
∵点P(b n ,b n+1 )在直线x-y+2上,∴b n -b n+1 +2=0,
∴b n+1 -b n =2,
∴数列{b n }是以b 1 =1为首项,2为公差的等差数列.
∴b n =1+(n-1)×2=2n-1.
(2)由(1)可得:a n b n =(2n-1)•2 n .
∴ T n =1×2+3× 2 2 + …+(2n-1)•2 n ,
2T n =1×2 2 +3×2 3 +…+(2n-3)•2 n +(2n-1)•2 n+1 ,
∴ - T n =1×2+2 ×2 2 +2× 2 3 +…+2×2 n -(2n-1)•2 n+1
=2×(2+2 2 +…+2 n )-2-(2n-1)•2 n+1
= 2×
2( 2 n -1)
2-1 -2-(2n-1)•2 n+1
=2 n+2 -6-(2n-1)•2 n+1
=(3-2n)•2 n+1 -6,
∴ T n =(2n-3)• 2 n+1 +6 .
1年前
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