Question

在数列an中,a1=1,an+1+an=6n.则a17=

Answer 1

an+1=-an+6n
令an+1+t=-(an+t)
∴t=-3n
∴an+1-3n=-(an-3n)
令an-3n=bn
∴{bn}是以-2为首项以-1为公比的等比数列
bn=2(-1)的n次方
an=2(-1)*n+3n
∴a17=49

呃、自己做的、不知道对不对~

Answer 2

a(n+1)+an=6n
a(n+1) + k1(n+1) + k2 = -(an+k1n+k2)
-2k1=6
k1=-3
-2k2-k1=0
-2k2+3=0
k2= 3/2

a(n+1)+an=6n
a(n+1) -3(n+1) +3/2 = -(an -3n + 3/2)
[a(n+1) -3(n+1) +3/2]/(an -3n + 3/2) = -1
(an -3n + 3/2)/(a1-3/2) = (-1)^(n-1)
an -3n + 3/2 = (1/2) (-1)^n
an = 3n -3/2 + (1/2) (-1)^n
a17 = 51-3/2-1/2
=49