Question

已知数列an满足a1=1,a2=2,an+1=an+an-1/2

Answer 1

an+2=(an+1)^2/（an+an+1）
2
边取倒数
1/a(n+2)=[an+a(n+1)]/[a(n+1)*a(n+1)]
a(n+1)/a(n+2)=[an+a(n+1)]/a(n+1)
=
an/a(n+1)
+
1
设bn=an/a(n+1)
则
b(n+1)=a(n+1)/a(n+2)
b(n+1)=bn+1
b(n+1)-bn=1
==>
bn
即{an/a(n+1)}
为等差数列
,首项为
b1=a1/a2=2
d=1
bn
=
an
/
a(n+1)
=
b1
+
(n-1)
d
=
2
+(n-1)
=
n+1
an/a(n+1)
=
n+1
a(n-1)/an
=
n
a(n-2)/a(n-1)=
n-1
...
a2/穿讥扁客壮九憋循铂末a3
=
3
a1/a2
=
2
两边相乘
a1/a(n+1)
=
2*3*4*5...*(n+1)
=(n+1)!
a(n+1)=a1/(n+1)!=1/(n+1)!
==>
an=1/[n!]

Answer 2

a(1)=1,a(2)=2,
a(n+2)=[a(n+1)+a(n)]/2,
a(n+2)-a(n+1)=[a(n)-a(n+1)]/2=(-1/2)[a(n+1)-a(n)],
b(n+1)=a(n+2)-a(n+1)=(-1/2)[a(n+1)-a(n)]=(-1/2)b(n),
{b(n)=a(n+1)-a(n)}是首项为a(2)-a(1)=1,公比为(-1/2)的等比数列。
b(n)=a(n+1)-a(n)=(-1/2)^(n-1),
a(n+1)
=
a(n)
+
(-1/2)^(n-1),
(-2)^na(n+1)
=
(-2)(-2)^(n-1)a(n)
-
2,
c(n)
=
(-2)^(n-1)a(n),
c(n+1)=(-2)^na(n+1)=(-2)(-2)^(n-1)a(n)
-
2
=
-2c(n)-2,
c(n+1)+2/3
=
-2c(n)-2+2/3
=
-2c(n)
-
4/3
=
(-2)[c(n)
+
2/3],
{c(n)+2/3}是首项为c(1)+2/3=a(1)+2/3=5/3,公比为（-2）的等比数列。
c(n)+2/3
=
(5/3)(-2)^(n-1),
c(n)
=
(5/3)(-2)^(n-1)
-
2/3
=
(-2)^(n-1)a(n),
a(n)
=
5/3
-
(2/3)(-1/2)^(n-1)