已知数列an是公比大于一的等比数列对任意的n属于有a(n+1)=a1+a2+...+a(n-1)+5/2an+1/2(1)求数列an的通项公式(2)设数列bn满足1/n(log3a1+log3a2+.+log3an+log3t)(n属于正整数)若bn为等差数列,求实数t的
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已知数列an是公比大于一的等比数列对任意的n属于有a(n+1)=a1+a2+...+a(n-1)+5/2an+1/2(1)求数列an的通项公式(2)设数列bn满足1/n(log3a1+log3a2+.+log3an+log3t)(n属于正整数)若bn为等差数列,求实数t的
已知数列an是公比大于一的等比数列对任意的n属于有a(n+1)=a1+a2+...+a(n-1)+5/2an+1/2
(1)求数列an的通项公式
(2)设数列bn满足1/n(log3a1+log3a2+.+log3an+log3t)(n属于正整数)
若bn为等差数列,求实数t的值及数列bn的通项公式
已知数列an是公比大于一的等比数列对任意的n属于有a(n+1)=a1+a2+...+a(n-1)+5/2an+1/2(1)求数列an的通项公式(2)设数列bn满足1/n(log3a1+log3a2+.+log3an+log3t)(n属于正整数)若bn为等差数列,求实数t的
1)
an= a1.q^(n-1)
a(n+1)=a1+a2+...+a(n-1)+(5/2)an+1/2 (1)
an=a1+a2+...+a(n-2)+(5/2)a(n-1)+1/2 (2)
(1)-(2)
a(n+1) - an = a(n-1) + (5/2)an - (5/2)a(n-1)
a(n+1) = (7/2)an - (3/2)a(n-1)
a1.q^n = a1.(7/2)q^(n-1) - (3/2)a1q^(n-2)
2q^2-7q+3 =0
(2q-1)(q-3)=0
q=3
from (1)
n=2
a3= a1+(5/2)a2 + 1/2
9a1 = a1 +(15a1/2) +1/2
a1/2 =1/2
a1 = 1
an = 3^(n-1)
(2)bn-bn-1=1/2+log3t/n+log3t/(n+1)
因为bn为等差数列
所以log3t=0
得t=1
bn=1/n(1+2+...+n+log3t)=1/n(1+2+...+n)=n-1/2
(1)
an= a1.q^(n-1)
a(n+1)=a1+a2+...+a(n-1)+(5/2)an+1/2 (1)
an=a1+a2+...+a(n-2)+(5/2)a(n-1)+1/2 (2)
(1)-(2)
a(n+1) - an =...
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(1)
an= a1.q^(n-1)
a(n+1)=a1+a2+...+a(n-1)+(5/2)an+1/2 (1)
an=a1+a2+...+a(n-2)+(5/2)a(n-1)+1/2 (2)
(1)-(2)
a(n+1) - an = a(n-1) + (5/2)an - (5/2)a(n-1)
a(n+1) = (7/2)an - (3/2)a(n-1)
a1.q^n = a1. (7/2)q^(n-1) - (3/2)a1q^(n-2)
2q^2-7q+3 =0
(2q-1)(q-3)=0
q=3
from (1)
n=2
a3= a1+(5/2)a2 + 1/2
9a1 = a1 +(15a1/2) +1/2
a1/2 =1/2
a1 = 1
an = 3^(n-1)
(2)
设数列bn满足1/n(log3a1+log3a2+。。+log3an+log3t)(n属于正整数)
bn 的定义不清楚!!!!!!!!!!!!
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