求一数列.高2. a(n+1)=2an/2an +1已知a1=1 a(n+1)=2an/2an+1求数列an的通项公式b(n+1)-d=1/2(bn-d)b(n+1)=1/2bn-1/2d对照系数-1/2d=1 d=-2 你这里是不是变错了?b(n+1)-d=1/2(bn-d)b(n+1)-2=1/2(bn-2)b(n)-2是以1/2为公比 b1-2=-1
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![求一数列.高2. a(n+1)=2an/2an +1已知a1=1 a(n+1)=2an/2an+1求数列an的通项公式b(n+1)-d=1/2(bn-d)b(n+1)=1/2bn-1/2d对照系数-1/2d=1 d=-2 你这里是不是变错了?b(n+1)-d=1/2(bn-d)b(n+1)-2=1/2(bn-2)b(n)-2是以1/2为公比 b1-2=-1](/uploads/image/z/14437983-39-3.jpg?t=%E6%B1%82%E4%B8%80%E6%95%B0%E5%88%97.%E9%AB%982.+a%28n%2B1%29%3D2an%2F2an+%2B1%E5%B7%B2%E7%9F%A5a1%3D1+a%28n%2B1%29%3D2an%2F2an%2B1%E6%B1%82%E6%95%B0%E5%88%97an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8Fb%28n%2B1%29-d%3D1%2F2%28bn-d%29b%28n%2B1%29%3D1%2F2bn-1%2F2d%E5%AF%B9%E7%85%A7%E7%B3%BB%E6%95%B0-1%2F2d%3D1+d%3D-2++%E4%BD%A0%E8%BF%99%E9%87%8C%E6%98%AF%E4%B8%8D%E6%98%AF%E5%8F%98%E9%94%99%E4%BA%86%EF%BC%9Fb%28n%2B1%29-d%3D1%2F2%28bn-d%29b%28n%2B1%29-2%3D1%2F2%28bn-2%29b%28n%29-2%E6%98%AF%E4%BB%A51%2F2%E4%B8%BA%E5%85%AC%E6%AF%94+b1-2%3D-1)
求一数列.高2. a(n+1)=2an/2an +1已知a1=1 a(n+1)=2an/2an+1求数列an的通项公式b(n+1)-d=1/2(bn-d)b(n+1)=1/2bn-1/2d对照系数-1/2d=1 d=-2 你这里是不是变错了?b(n+1)-d=1/2(bn-d)b(n+1)-2=1/2(bn-2)b(n)-2是以1/2为公比 b1-2=-1
求一数列.高2. a(n+1)=2an/2an +1
已知a1=1
a(n+1)=2an/2an+1
求数列an的通项公式
b(n+1)-d=1/2(bn-d)
b(n+1)=1/2bn-1/2d
对照系数
-1/2d=1 d=-2 你这里是不是变错了?
b(n+1)-d=1/2(bn-d)
b(n+1)-2=1/2(bn-2)
b(n)-2是以1/2为公比 b1-2=-1为首项的等比数列
谁和你说bn-2是等比数列?
求一数列.高2. a(n+1)=2an/2an +1已知a1=1 a(n+1)=2an/2an+1求数列an的通项公式b(n+1)-d=1/2(bn-d)b(n+1)=1/2bn-1/2d对照系数-1/2d=1 d=-2 你这里是不是变错了?b(n+1)-d=1/2(bn-d)b(n+1)-2=1/2(bn-2)b(n)-2是以1/2为公比 b1-2=-1
两边取倒数
1/a(n+1)=(2an+1)/2an=1+1/2an
令bn=1/an
b1=1/a1=1
则b(n+1)=1/2bn+1
用待定系数法
b(n+1)-d=1/2(bn-d)
b(n+1)=1/2bn+1/2d
对照系数
-1/2d=1 d=2
b(n+1)-2=1/2(bn-2)
b(n)-2是以1/2为公比 b1-2=-1为首项的等比数列
则bn-2=(1/2)^(n-1)(b1-2)=-(1/2)^(n-1)
所以bn=-(1/2)^(n-1)+2
所以an=1/bn=1/[-(1/2)^(n-1)+2]
yg
1/a[n+1]=2an+1/2an 1/a[n+1]=1+1/2an
设数列 bn=1/an+2
由 1/a[n+1]=1+1/2an得1/an=2/a[n+1]-2
b[n+1]=1/a[n+1]-2 bn=2/a[n+1]-4
b[n+1]/bn=1/2 所以bn为公差为1/2的等比数列
因为 bn=1/an+2 a1=1...
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1/a[n+1]=2an+1/2an 1/a[n+1]=1+1/2an
设数列 bn=1/an+2
由 1/a[n+1]=1+1/2an得1/an=2/a[n+1]-2
b[n+1]=1/a[n+1]-2 bn=2/a[n+1]-4
b[n+1]/bn=1/2 所以bn为公差为1/2的等比数列
因为 bn=1/an+2 a1=1 所以b1=-1
bn=-1/[2的n-1次幂]
1/an=-1/[2的n-1次幂]-2 所以an=2的n-1幂/[2的n次幂-1]
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