问题
解答题
数列{an}满足a1=
(1)写出a2,a3,a4; (2)猜出an的表达式,并用数学归纳法证明. |
答案
(1)令n=2,∵a1=
,∴S2=1 6
a2,即a1+a2=3a2.∴a2=2×(2+1) 2
.1 12
令n=3,得S3=
a3,即a1+a2+a3=6a3,∴a3=3×(3+1) 2
.1 20
令n=4,得S4=
a4,a1+a2+a3+a4=10a4,∴a4=4×(4+1) 2
.1 30
(2)猜想an=
,下面用数学归纳法给出证明.1 (n+1)(n+2)
①当n=1时,a1=
=1 6
结论成立.1 (1+1)(1+2)
②假设当n=k时,结论成立,即ak=
,1 (k+1)(k+2)
则当n=k+1时,Sk=
ak=k(k+1) 2
•k(k+1) 2 1 (k+1)(k+2)
=
,Sk+1=k 2(k+2)
ak+1,(k+1)(k+2) 2
即Sk+ak+1=
ak+1.(k+1)(k+2) 2
∴
+ak+1=k 2(k+2)
ak+1.(k+1)(k+2) 2
∴ak+1=
=k 2(k+2)
-1(k+1)(k+2) 2
=k k(k+3)(k+2)
.1 (k+2)(k+3)
∴当n=k+1时结论成立.
由①②可知,对一切n∈N+都有an=
成立.1 (n+1)(n+2)