Question

设数列{an}满足关系an=3/2（an-1）+5（n≥2）,a1=-17/2.令bn=an+10,求数列{bn}的前n

Answer 1

an-3/2(an-1) =5 (1)式

(an-1) - 3/2(an-2) =5 (2)式 .

(an-2) - 3/2(an-3) =5 (3)式 .

.

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a2- 3/2(a1) =5 (n-1)式

上述(2)式乘以3/2,上述(3)式乘以(3/2)^2,.上述(n-1)式乘以(3/2)^n-2,将全部n-1个等式相加,就得到：

an - a1(3/2)^n-1 = 5+ 5x(3/2) + 5x(3/2)^2 + .+ 5x(3/2)^n-2(本步骤中x代表乘号)

等式右侧是个首项5,公比3/2的等比数列的n-1项求和,所以它等于10(3/2)n-1 - 10

即：an - a1(3/2)^n-1 = 10(3/2)n-1 - 10

将 a1= -17/2带入,得到an = (3/2)^n -10

所以bn = (3/2)^n; bn是个首项3/2,公比3/2的等比数列

Sn = 3(3/2)^n -3