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
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
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