已知数列满足a1=1,a(n+1)=2an+1 (Ⅰ)求数列的通项公式; (Ⅱ)若数列满足,4^b1-1*4^b2-1*……4^bn-1=(a
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![已知数列满足a1=1,a(n+1)=2an+1 (Ⅰ)求数列的通项公式; (Ⅱ)若数列满足,4^b1-1*4^b2-1*……4^bn-1=(a](/uploads/image/z/10338080-32-0.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%E6%BB%A1%E8%B6%B3a1%3D1%2Ca%28n%2B1%29%3D2an%2B1+%EF%BC%88%E2%85%A0%EF%BC%89%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%9B+%EF%BC%88%E2%85%A1%EF%BC%89%E8%8B%A5%E6%95%B0%E5%88%97%E6%BB%A1%E8%B6%B3%2C4%5Eb1-1%2A4%5Eb2-1%2A%E2%80%A6%E2%80%A64%5Ebn-1%3D%28a)
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已知数列满足a1=1,a(n+1)=2an+1 (Ⅰ)求数列的通项公式; (Ⅱ)若数列满足,4^b1-1*4^b2-1*……4^bn-1=(a
已知数列满足a1=1,a(n+1)=2an+1 (Ⅰ)求数列的通项公式; (Ⅱ)若数列满足,4^b1-1*4^b2-1*……4^bn-1=(a
已知数列满足a1=1,a(n+1)=2an+1 (Ⅰ)求数列的通项公式; (Ⅱ)若数列满足,4^b1-1*4^b2-1*……4^bn-1=(a
(1)我告诉你一个绝招:::叫特征根法:以这个题为例你自己在研究一下(最好去求一下斐波拉契数列:a1=a2=1,a(n+2)=a(n+1)+a(n)).方法是:设:r-2=0,得:r=2,于是设:a(n)=k2^n+p,
而:a1=1,a2=3,代入:
2k+p=1
4k+p=3
得:k=1;p=-1
于是:a(n)=2^n-1
(2)其方法,也是很简单的,由于时间关系.这个你先自己解决.
b(n)+2是首项为4,公比为2的等比数列 b(n)+2=2^(n+1) b(n)=2∵a1=2,a2=4,b(n)=a(n+1)-a(n) ∴b1=a2-a1=2 ∵b(
第一问、方法:a(n+1)+1=2(an+1),另c(n+1)=a(n+1)+1,则c(n)是以首项为2公比也为2的等比数列
第二问,方法:你没写完吧??????
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