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问题: 数列问题,求an

数列an,若a1,a2-a1,a3-a2 ……an-a(n-1)是首项为1,公比为1/3的等比数列,则an=

解答:

a1 = 1
a2 - a1 = (1/3)
a3 - a2 = (1/3)^2
a4 - a3 = (1/3)^3
.................
an - a(n-1) = (1/3)^(n-1)

相加得
an = 1 + (1/3) + (1/3)^2 + (1/3)^3 + ... + (1/3)^(n-1)
  = 1 * [1 - (1/3)^n] / (1 - 1/3)
  = (3/2)(1 - 1/3^n)