Question

设数列An的前n项和为Sn,已知A1=A2=1,Bn=nSn+(n+2)An,数列Bn是公差为d的等差数列,求d的值.求数列An的通项

求数列An的通项公式

Answer 1

bn = b1+(n-1)d
Sn = a1=a2+...+an

bn =nSn+(n+2)an
n=1
b1= a1+3a1
= 4
n= 2
b2= 2(a1+a2) + 4a2
= 2(2)+4
=8
d= b2-b1= 4
bn = b1+(n-1)d = 4 +(n-1)4 = 4n

bn =nSn+(n+2)an
4n = nSn +(n+2)an
= nSn +(n+2)[Sn-S(n-1)]
(2n+2)Sn = (n+2)S(n-1) + 4n
2Sn/(n+2) = S(n-1)/(n+1) + 4n/[(n+1)(n+2)]
2[Sn/(n+2) - 4/(n+2)] = S(n-1)/(n+1) - 4/(n+1)
[(Sn/(n+2) - 4/(n+2)]/[ S(n-1)/(n+1) - 4/(n+1)] = 1/2
[Sn/(n+2) - 4/(n+2)]/[ (S1)/3 - 4/3] = (1/2)^(n-1)
Sn/(n+2) - 4/(n+2) = - (1/2)^(n-1)
Sn = [4/(n+2) -(1/2)^(n-1)] (n+2)
= 4 - (n+2)(1/2)^(n-1)
an = Sn-S(n-1)
=-(n+2) (1/2)^(n-1) + (n+1) (1/2)^(n-2)
=-(1/2)^(n-1). [ n+2 - 2(n+1)]
= n.(1/2)^(n-1)

Answer 2

因为Bn=nSn+(n+2)An
所以Bn-1=（n-1）Sn-1+(n+1)An-1；B1=4
d=Sn+An-An-1
A1=A2=1
d=S2=2
所以Bn=4+2（n-1）=2(n+1)