数列{an}中,a1=0,a2=2,a(n+2)-6a(n+1)+5an=2^n,求an
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数列{an}中,a1=0,a2=2,a(n+2)-6a(n+1)+5an=2^n,求an
(n+2),(n+1)均为下标
(n+2),(n+1)均为下标
![数列{an}中,a1=0,a2=2,a(n+2)-6a(n+1)+5an=2^n,求an](/uploads/image/z/16540403-59-3.jpg?t=%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%2Ca1%3D0%2Ca2%3D2%2Ca%28n%2B2%29-6a%28n%2B1%29%2B5an%3D2%5En%2C%E6%B1%82an)
1、设b[n]=a[n]+⅓×2^n,那么
b[1] = a[1] + ⅓×2 = ⅔
b[2] = a[2] + ⅓×2^2 = 10/3
且有 a[n] = b[n] - ⅓×2^n,代入 a[n+2] - 6a[n+1] + 5a[n] = 2^n 并整理,有
b[n+2] - 6b[n+1] + 5b[n] = 0
2、数列 b[n] 的通项公式可以由特征方程给出
b[n] = 5^n × 2/15
或者这样得到:
由b[n+2] - 6b[n+1] + 5b[n] = 0有 b[n+2] - b[n+1] = 5(b[n+1] - b[n]),所以
c[n] = b[n+1] - b[n] 是比例为5的等比数列,且 c[1] = b[2] - b[1] = 8/3,
所以c[n] = 8/3 × 5^(n-1),从而
b[n] = c[n-1] + b[n-1] = c[n-1] + c[n-2] + b[n-2] = ...= c[n-1] + c[n-2] + c[n-3] + ...+c[1] + b[1]
=5^n × 2/15
3、由b[n]的通项公式,可以得到a[n]的通项公式为
a[n] = 5^n × 2/15 - ⅓×2^n
希望能解决您的问题.
b[1] = a[1] + ⅓×2 = ⅔
b[2] = a[2] + ⅓×2^2 = 10/3
且有 a[n] = b[n] - ⅓×2^n,代入 a[n+2] - 6a[n+1] + 5a[n] = 2^n 并整理,有
b[n+2] - 6b[n+1] + 5b[n] = 0
2、数列 b[n] 的通项公式可以由特征方程给出
b[n] = 5^n × 2/15
或者这样得到:
由b[n+2] - 6b[n+1] + 5b[n] = 0有 b[n+2] - b[n+1] = 5(b[n+1] - b[n]),所以
c[n] = b[n+1] - b[n] 是比例为5的等比数列,且 c[1] = b[2] - b[1] = 8/3,
所以c[n] = 8/3 × 5^(n-1),从而
b[n] = c[n-1] + b[n-1] = c[n-1] + c[n-2] + b[n-2] = ...= c[n-1] + c[n-2] + c[n-3] + ...+c[1] + b[1]
=5^n × 2/15
3、由b[n]的通项公式,可以得到a[n]的通项公式为
a[n] = 5^n × 2/15 - ⅓×2^n
希望能解决您的问题.
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