若数列{a}满足关系a1=2,a=3a+2,求数列的通项公式(我的思路:由已知,得a=3a+2,a/a=3,即{a}是等比数列,a1=2,q=3,所以通项公式为2×3^(n-1).请问这思路错在哪里)
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![若数列{a}满足关系a1=2,a=3a+2,求数列的通项公式(我的思路:由已知,得a=3a+2,a/a=3,即{a}是等比数列,a1=2,q=3,所以通项公式为2×3^(n-1).请问这思路错在哪里)](/uploads/image/z/2768806-46-6.jpg?t=%E8%8B%A5%E6%95%B0%E5%88%97%7Ba%7D%E6%BB%A1%E8%B6%B3%E5%85%B3%E7%B3%BBa1%3D2%2Ca%3D3a%2B2%2C%E6%B1%82%E6%95%B0%E5%88%97%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%88%E6%88%91%E7%9A%84%E6%80%9D%E8%B7%AF%EF%BC%9A%E7%94%B1%E5%B7%B2%E7%9F%A5%2C%E5%BE%97a%3D3a%2B2%2Ca%2Fa%3D3%2C%E5%8D%B3%7Ba%7D%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2Ca1%3D2%2Cq%3D3%2C%E6%89%80%E4%BB%A5%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E4%B8%BA2%C3%973%5E%28n-1%29.%E8%AF%B7%E9%97%AE%E8%BF%99%E6%80%9D%E8%B7%AF%E9%94%99%E5%9C%A8%E5%93%AA%E9%87%8C%EF%BC%89)
若数列{a}满足关系a1=2,a=3a+2,求数列的通项公式(我的思路:由已知,得a=3a+2,a/a=3,即{a}是等比数列,a1=2,q=3,所以通项公式为2×3^(n-1).请问这思路错在哪里)
若数列{a
(我的思路:由已知,得a
若数列{a}满足关系a1=2,a=3a+2,求数列的通项公式(我的思路:由已知,得a=3a+2,a/a=3,即{a}是等比数列,a1=2,q=3,所以通项公式为2×3^(n-1).请问这思路错在哪里)
由a
这种题做熟了发现解法都是固定的,就是想办法变成等比数列求解(一般是an加一个常数)
假设a(n+1)+ k =3(an + k),解出 k = 1.说明{an + 1}是等比数列,通项为
an+1 = 3*3^(n-1) = 3^n
所以an = 3^n -1
a
a
令b
b
b
故a
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