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帮忙·········帮忙·········帮忙·········a(n)=aq^(n-1),8/3^n=a(n+1)+a(n)=aq^n+aq^(n-1)=(q+1)aq^(n-1)=(8/3)(1

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a(n) = aq^(n-1),
8/3^n = a(n+1) + a(n) = aq^n + aq^(n-1) = (q+1)aq^(n-1) = (8/3)(1/3)^(n-1),
q = 1/3,8/3 = (q+1)a = (4/3)a,a= 2.
a(n) = 2/3^(n-1).
a(n) - a(n+1) = 2/3^(n-1) - 2/3^n = 4/3^n
b(n) = log_{3}[a(n)-a(n+1)] = log_{3}[4/3^n] = log_{3}(4) - n = log_{3}(4) - 1 + (n-1)(-1),
{b(n)}为首项为log_{3}(4) - 1,公差为(-1)的等差数列.
s(n) = 2[1 - 1/3^n]/(1-1/3) = 3[1 - 1/3^n] = 3 - 1/3^(n-1),
a(n)s(n) = 6/3^(n-1) - 2/9^(n-1),
t(n) = 6[1 - 1/3^n]/(1-1/3) - 2[1-1/9^n]/(1-1/9)
= 9[1- 1/3^n] - (9/4)[ 1 - 1/9^n]
= 9 - 9/3^n - 9/4 + (9/4)(1/9^n)
= 27/4 - 1/3^(n-2) + (1/4)[1/9^(n-1)]

这图有点模糊,等好心芝麻吧
亲,我走了