高二数学 数列 求步骤
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高二数学数列题！！！要步骤！！！~
a(n+1)=8an-6bn (1)
b(n+1)=6an-4bn (2)
(1)-(2)
a(n+1)-b(n+1) =2(an -bn)
=>{an -bn}是等比数列, q=2
an -bn = 2^(n-1).(a1 -b1)
=2^n
bn = an - 2^n (3)
sub (3) into (1)
a(n+1) = 8an - 6[an - 2^n]
a(n+1)/2^(n+1) -an/2^n = 3
=>{an/2^n} 是等差数列, d=3
an/2^n - a1/2 = 3(n-1)
an/2^n = (6n -5)/2
an =(6n -5). 2^(n-1)
from (3)
bn = an - 2^n
=(6n -5). 2^(n-1) - 2^n
=(6n -7). 2^(n-1)
5年=60月
公差d=2元
a1=2(1-1)+500=500
Sn=a1*n+n(n-1)d/2
S60=500*60+60*59*2/2=33540元
总消耗500000+33540=533540元
每天消耗：
533540÷5÷365≈292.35元
如果那个常数角标为k+2的话那容易，根据求根公式，然后ak和ak+2用ak+1表示 ，化简就可以得到一个常数-1，另一个为-ak+2/ak.1/(1-ak+2/ak)化简得到一个常数与ak的乘积，ak为等差数列，常数与ak的乘积也为等差数列。
a1=1,b1=-1a(n+1)=8an-6bn (1)
b(n+1)=6an-4bn (2)
(1)-(2)
a(n+1)-b(n+1) =2(an -bn)
=>{an -bn}是等比数列, q=2
an -bn = 2^(n-1).(a1 -b1)
=2^n
bn = an - 2^n (3)
sub (3) into (1)
a(n+1) = 8an - 6[an - 2^n]
a(n+1)/2^(n+1) -an/2^n = 3
=>{an/2^n} 是等差数列, d=3
an/2^n - a1/2 = 3(n-1)
an/2^n = (6n -5)/2
an =(6n -5). 2^(n-1)
from (3)
bn = an - 2^n
=(6n -5). 2^(n-1) - 2^n
=(6n -7). 2^(n-1)
