Question

高一数列题

设数列an,Sn是它的前n项和,并且S(n+1)=4an+2 (n属于N*) a1=1
(1)设bn=a(n+1)-2an,求证数列bn是等比数列
(2)设cn=an/2^n，求证cn是等差数列
(3)求数列an的通项公式及前n项和公式

Answer 1

S(n+1)=4an+2
sn=4a(n-1)+2
S(n+1)-sn=4an+2-[4a(n-1)+2]
a(n+1)=4an-4a(n-1)
a(n+1)-2an=2[an-2a(n-1)]
[a(n+1)-2an]/[an-2a(n-1)]=2
(1)bn=a(n+1)-2an是等比数列
a(n+1)-2an=b1*2^n=3*2^n
a(n+1)/2^(n+1)-an/2^n=3/2
(2)cn=an/2^n是等差数列
an/2^n=c1+(n-1)3/2=[3n-2]/2
an=[3n-2]*2^n/2
sn=1/2[1*2^1+4*2^2+7*2^3+……+(3n-2)*2^n]
2sn=1*2^1+4*2^2+7*2^3+……+(3n-2)*2^n
4sn=1*2^2+4*2^3+7*2^4+……+(3n-2)*2^(n+1)
-2sn=1*2^1+3*2^2+3*2^3+3*2^4+……+3*2^n-(3n-2)*2^(n+1)
=1*2^1+3[2^2+2^3+2^4+……+2^n]-(3n-2)*2^(n+1)
=1*2^1+3[2^(n+1)-2^2]-(3n-2)*2^(n+1)
sn=[(3n-2)*2^(n+1)-1*2^1-3[2^(n+1)-2^2]]/2=ok

Answer 2

（1) 、解：a(n+2)=S(n+2)-S(n+1)
=4a(n+1)+2-4an-2
=4(a(n+1)-an)(n>=1)
b(n+1)/bn=a(n+2)-2a(n+1)/a(n+1)-2an
=4(a(n+1)-an)-2a(n+1)/a(n+1)-2an
=2a(n+1)-4an/a(n+1)-2an
=2 (n>=1)
b1=a2-2a1
=5-2
=3
所以bn是以3为首项、2为公比的等比数列。
(2)、解：Cn=an/2^n
C(n+1)-Cn=a(n+1)/2^(n+1)-an/2^n
=a(n+1)-2an/2^(n+1)
=bn/2^(n+1)
=3*2^（n-1）/2^(n+1)
=3/4(n>=1)
所以C1是以1/2为首项、3/4为公差的等差数列。
（3）解：Cn=an/2^n
an=2^n*Cn
=2^n*(1/2+(n-1)3/4)
=2^n*((3n-1)/4)
Sn=2*(2/4)+2^2*( 5/4)+2^3( 8/4)+...+...+ 2^n*((3n-1)/4)
2Sn= 2^2*(2/4)+2^3*( 5/4)+2^4( 8/4)+...+...+2^(n-1)*((3n-4)/4)+ 2^(n+1)*((3n-1)/4)
Sn=2Sn-Sn
=-1-2^2*(3/4)-2^3*(3/4)-...-...-2^n*(3/4)+ 2^(n+1)*((3n-1)/4)
=-1-3/4(2^(n-1)-1) )+ 2^(n+1)*((3n-1)/4)
=-1+3/4-2^(n+1)*3/16 + 2^(n+1)*((3n-1)/4)
= 2^(n+1)*(12n-15)/16+1/4
好了，额好久没计算过了。弃武从文了。有不足之处请见谅。第三题可能会有误。