Question

数列{an}前n项和为sn，sn=2an-3n（n属于n+），求数列{an}的通项公式，在线等

Answer 1

Sn=2an-3n

a1=S1=2a1-3
a1=3
n>=2:
an=Sn-S(n-1)=2an-3n-2a(n-1)+3(n-1)
an=2a(n-1)+3
an+3=2[a(n-1)+3]
故数列{an+3}是以a1+3=6,公比是2的等比数列.
即an+3=6*2^(n-1)
an=3*2^n-3
a1=3*2-3=3,符合.
故an=3*2^n-3

Answer 2

先解a1：a1 = S1 = 2a1-3，a1 = 3
an = Sn-S(n-1) = (2an-3n)-(2a(n-1)-3(n-1) ) = 2an-2a(n-1)-3
移项：an = 2a(n-1)+3，an+3 = 2a(n-1)+6，an+3 = 2( a(n-1)+3 )
令Bn = an+3，Bn = 2B(n-1)，B1 = a1+3 = 6，Bn = 6 * 2 ^ (n-1) = 3 * 2 ^ n
所以 an = Bn - 3 = 3 * 2 ^ n - 3 = 3 * (2 ^ n - 1)

追问

a

Answer 3

an=sn-sn-1=2an-3n-(2an-1-3(n-1) 得到an=2an-1-3 s1=a1=2a1-3 得到a1=3 a2=0 a3=-3所以 an=2an-1-3(n.>1且属于整数）