Question

（Ⅰ）在数列{an}中,a1=1,a(n+1)=6n-an,求an ；（Ⅱ）在数列{an}中,a1=1,

Answer 1

1

an+1=an+6n

an=an-1+6(n-1)

..

a2=a1+6*2

a1=6-5

Sn=Sn-1+6*(1+2+..+n)-5

an=6*(1+n)n/2-5=3n(n+1)-5

2

an*an+1=3^n

a1*a2=3,a2=3,a2*a3=3^2,a3=3 a3*a4=3^3,a4=3^2,a4*a5=3^4,a5=3^2

an是序数奇数（偶数）特征的等比数列

a2k=3^k,a2k+1=3^k

a2k*a2k+1=3^2k ,a2k+1*a2k+2=3^k*3^k+1=3^2k+1

3

a1=1

a2=a1-1=0,a3=3,a4=3+1=4,a5=4+9=13,a6=13-1=12

a2k=a2k-1+(-1)^k a2k+1=a2k+3^k

a2k-2=a2k-3+(-1)^(k-1) a2k-1=a2k-2+3^(k-1)

..

a3=

a2=a1+(-1) a3=a2+3

a1=1

n=4k,

Sn=Sn-1+3^1+3^2+..+3^(2k-1)=Sn-1+3*(1-3^(2k))/(1-3)=Sn-1+3(3^(2k)-1)/2

an=3[3^(2k)-1]/2

n=4k+2

Sn=Sn-1 -1+3^1+3^2+..+3^(2k+1)=Sn-1+3*(3^(2k+2)-1)/2

an=3[3^(2k+2) -1]/2 -1

n=4k+1

Sn=Sn-1+3^1+3^2+..+3^2k=Sn-1+3*(3^(2k+1)-1)/2

an=3*[3^(2k+1)-1]/2

n=4k+3

Sn=Sn-1 -1+3^1+3^2+..+3^(2k+2)=Sn-1+3*(3^(2k+3)-1)/2 -1

an=3*[3^(2k+3)-1]/2 -1