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已知等差数列的前n项和Sn=a•2^(n+1)+1/8,则a的值为____
请详细讲解
人气:367 ℃ 时间:2020-06-13 15:42:42
解答
Sn=a•2^(n+1)+1/8
n=1
a1 = 4a+1/8
n=2
a1+a2 = 8a +1/8
a2 = 4a
n=3
a1+a2+a3 = 16a+1/8
a3 = 8a
a3-a2 = a2-a1
8a-4a= 4a -(4a+1/8)
4a= -1/8
a = -1/32
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