(高中数学)已知数列{an}满足a1=3,an+1(n+1是脚标)=2an(只有n是脚标)+1,求数列的通项公式.我知道大概步骤,就是将式子化为an+1 +1=2(an+1).进而用累乘法求答案的.an +1 =2(an-1 +1),an-1 +1=2(an-2
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![(高中数学)已知数列{an}满足a1=3,an+1(n+1是脚标)=2an(只有n是脚标)+1,求数列的通项公式.我知道大概步骤,就是将式子化为an+1 +1=2(an+1).进而用累乘法求答案的.an +1 =2(an-1 +1),an-1 +1=2(an-2](/uploads/image/z/11715542-62-2.jpg?t=%EF%BC%88%E9%AB%98%E4%B8%AD%E6%95%B0%E5%AD%A6%EF%BC%89%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D3%2Can%2B1%EF%BC%88n%2B1%E6%98%AF%E8%84%9A%E6%A0%87%EF%BC%89%3D2an%EF%BC%88%E5%8F%AA%E6%9C%89n%E6%98%AF%E8%84%9A%E6%A0%87%EF%BC%89%2B1%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.%E6%88%91%E7%9F%A5%E9%81%93%E5%A4%A7%E6%A6%82%E6%AD%A5%E9%AA%A4%2C%E5%B0%B1%E6%98%AF%E5%B0%86%E5%BC%8F%E5%AD%90%E5%8C%96%E4%B8%BAan%2B1+%2B1%3D2%28an%2B1%29.%E8%BF%9B%E8%80%8C%E7%94%A8%E7%B4%AF%E4%B9%98%E6%B3%95%E6%B1%82%E7%AD%94%E6%A1%88%E7%9A%84.an+%2B1+%3D2%EF%BC%88an-1+%2B1%EF%BC%89%2Can-1+%2B1%3D2%EF%BC%88an-2)
(高中数学)已知数列{an}满足a1=3,an+1(n+1是脚标)=2an(只有n是脚标)+1,求数列的通项公式.我知道大概步骤,就是将式子化为an+1 +1=2(an+1).进而用累乘法求答案的.an +1 =2(an-1 +1),an-1 +1=2(an-2
(高中数学)已知数列{an}满足a1=3,an+1(n+1是脚标)=2an(只有n是脚标)+1,求数列的通项公式.
我知道大概步骤,就是将式子化为an+1 +1=2(an+1).进而用累乘法求答案的.
an +1 =2(an-1 +1),an-1 +1=2(an-2 +1)``````````````a2 +1=2(a1 +1)`
然后各等式相乘,从而约去一部分项,得到最后结果
但此时让我不解的是,那约去的这些项中,就没可能有为0的情况吗?如果有为0的,不就不能约去了吗?
这是怎么回事?希望明白的人解释下,
(高中数学)已知数列{an}满足a1=3,an+1(n+1是脚标)=2an(只有n是脚标)+1,求数列的通项公式.我知道大概步骤,就是将式子化为an+1 +1=2(an+1).进而用累乘法求答案的.an +1 =2(an-1 +1),an-1 +1=2(an-2
an = 2(a) +1
= 2[2(a) +1] +1
= 2^2 (a) + 2 + 1
= 2^2[2 (a) + 1] + 2 + 1
= 2^3 (a) + 2^2 + 2 + 1
= ……
= 2^(n-1) a1 + 2^(n-2) + 2^(n-3) +……+ 2 + 1
= 3 x 2^(n-1) + 2^ (n-1) - 1
= 4 x 2^(n-1) - 1
= 2^(n+1) - 1