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

来源:学生作业帮助网 编辑:作业帮 时间:2024/12/02 08:01:06
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
xRMN@J-mf²"&& Fbܨ( XB;oBh of:PDv4y{oƮVKvK/F+vrDG=0 zA|"YtYk/QRHsТmQv}TB{`0Y?ճ#9XՊ?5ފ幑Ѹx#,E euNZY 4 emz=O{x>Xpb0vMz6]+DNla8@|&)c1+q#"l! ;pNRA H F>sbVIحs9&#v],cr^84Yt릩B!—_M  'B `P+[?G!^& لz|W3{

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
∵a(n+1)=2an+1
∴a(n+1)+1=2an+2=2﹙an+1﹚
∴[a(n+1)+1]/﹙an+1﹚=2
∵a1+1=1+1=2
∴数列﹛an+1﹜是以2为首相,2为公比的等比数列
∴an+1=2×2^﹙n-1﹚=2^n
∴an=2^n-1

a(n+1)=2an+1
因为a(n+1)+1=2an+2=2﹙an+1﹚
所以a(n+1)+1]/﹙an+1﹚=2
因为a1+1=1+1=2
所以数列﹛an+1﹜是的首相是2.公差q=2
an+1=2×2^﹙n-1﹚=2^n
an=2^n-1

^(b1-1).4^(b2-1)...4^(bn-1) = (an+1)^bn
4^(b1+b2+..+bn- n) = (a(n+1))^bn
2^(2(b1+b2+..+bn- n)) = (2^n)^bn = 2^(nbn)
nbn = 2(b1+b2+..+bn- n)
b1+b2+..+bn = (nbn +2n)/2
= n(bn+2)/2
bn是等差数列,
b1=2