高二数学,求出每题的通项公式,有详细过程
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(4)
a1=2
a(n+1)= 2an +2^(n+1)
a(n+1)/2^(n+1) - an/2^n = 1
=> {an/2^n} 是等差数列, d=1
an/2^n -a1/2 = n-1
an/2^n = n
an =n.2^n
(5)
a1=2
a(n+1) = (1/2)an +1/2
a(n+1) - 1 = (1/2) [ an -1]
=> { an -1} 是等比数列, q=1/2
an -1 =(1/2)^(n-1). (a1 -1)
=(1/2)^(n-1)
an = 1+ (1/2)^(n-1)
(6)
a1=1
a(n+1) = an/(an+2)
1/a(n+1) = (an+2)/an
= 1+ 2/an
1/a(n+1) +1 =2[ 1/an +1]
=> {1/an +1}是等比数列, q=2
1/an +1= 2^(n-1) . (1/a1 +1)
= 2^n
an = 1/(2^n -1)
(7)
a1=3
an= a(n+1) +5an -a(n+1)
题目有误,查一下吧!
a1=2
a(n+1)= 2an +2^(n+1)
a(n+1)/2^(n+1) - an/2^n = 1
=> {an/2^n} 是等差数列, d=1
an/2^n -a1/2 = n-1
an/2^n = n
an =n.2^n
(5)
a1=2
a(n+1) = (1/2)an +1/2
a(n+1) - 1 = (1/2) [ an -1]
=> { an -1} 是等比数列, q=1/2
an -1 =(1/2)^(n-1). (a1 -1)
=(1/2)^(n-1)
an = 1+ (1/2)^(n-1)
(6)
a1=1
a(n+1) = an/(an+2)
1/a(n+1) = (an+2)/an
= 1+ 2/an
1/a(n+1) +1 =2[ 1/an +1]
=> {1/an +1}是等比数列, q=2
1/an +1= 2^(n-1) . (1/a1 +1)
= 2^n
an = 1/(2^n -1)
(7)
a1=3
an= a(n+1) +5an -a(n+1)
题目有误,查一下吧!
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