已知数列
|
S1=
=1 1•4
,S2=1 4
+1 1•4
=1 4•7
S3=2 7
+1 1•4
+1 4•7
=1 7•10
,S4=3 10
+1 1•4
+1 4•7
+1 7•10
=1 10•13
----------(4分)4 13
猜想:Sn=
----------------------------(6分)n 3n+1
证明:(1)当n=1 时,由上面计算知结论正确.
(2)假设n=k时等式成立,即Sk=
,k 3k+1
则当n=k+1时
Sk+1=Sk+
=1 (3k+1)(3k+4)
+k 3k+1 1 (3k+1)(3k+4) =
=3k2+4k+1 (3k+1)(3k+4)
=(3k+1)(k+1) (3k+1)(3k+4)
=k+1 3k+4 k+1 3(k+1)+1
即n=k+1时等式成立
由(1),(2)知,等式对任意正整数都成立-------------------------(14分)