(2)
已知数列
可推出A
n
=
7
(3)
由通项公式a
n
=1-3n,可知
a
1
=1-3=-2
a
2
=-5=a
1
-3
a
3
=-8=a
2-3
……
a
n
=a
n-1
-3
(6)
S
n
=2n
2
-3n,
n=1时,S
1=
a
1
=-1;
由
由1-2
得,a
n
=4n-5;
则a
1
=-1
a
2
=7
a
3
=15
….
a
n
=A
n-1
+8,
a
n
=(a
n
-a
n-1
)+(a
n-1
-a
n-2
)+(a
n-3
-a
n-2
)+…..+a
1
=8+8+….+(-1)
=8(n-1)-1=8n-9