在等差数列an中 a3=13 a6=19 求an的通项公式
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发布时间：2023-12-23 19:01
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热心网友
时间：2024-03-08 17:45
a3*a6 = (a4 -d)(a4 + 2d) = (2 -d)(2 + 2d) = 4 +2d -d² = -8
d² - 2d -6 = 0
(d - 3)(d+2) = 0
d = 3:此时d2 = d4 -2d = 2 -2*3 = -4 < 0,舍去
d = -2 (此时d2 = d4 -2d = 2 -2*(-2) = 6 > 0,符合题意)
a1 = a4 - 3d = 2-3(-2) = 8
an = 8 -2(n-1) = 10 -2n
2.
bn = (√2)^(an) = (√2)^(10 -2n) =2^(5 -n)=32*2^(-n)
Sn = 32[2^(-1) + 2^(-2) + ...+ 2^(-n)]
括号内为首项为1/2,公比为1/2的等比数列前n项和,= [2^(-1) - 2^(n+1)]/(1 - 1/2) = 1 - 2^(-n)
Sn = 32[1 - 2^(-n)]
