Question

已知數(shù)列

8•1

12•32

，  

8•2

32•52

， …， 

8n

(2n-1)2(2n+1)2

， ….S_n為其前n項(xiàng)和．計(jì)算得S1=

8

9

，  S2=

24

25

，  S3=

48

49

，  S4=

80

81

.觀察上述結(jié)果，推測出計(jì)算S_n的公式，并用數(shù)學(xué)歸納法加以證明．

Answer 1

觀察分析題設(shè)條件可知Sn=

(2n+1)2-1

(2n+1)2

(n∈N)
證明如下：（1）當(dāng)n=1時(shí)，S1=

32-1

32

=

8

9

，等式成立．
（Ⅱ）設(shè)當(dāng)n=k時(shí)等式成立，即Sk=

(2k+1)2-1

(2k+1)2

.則Sk+1=Sk+

8(k+1)

(2k+1)2(2k+3)2

=

(2k+1)2-1

(2k+1)2

+

8(k+1)

(2k+1)2(2k+3)2

=

[(2k+1)2-1](2k+3)2+8(k+1)

(2k+1)2(2k+3)2

=

(2k+1)2(2k+3)2-(2k+3)2+8(k+1)

(2k+1)2(2k+3)2

=

(2k+1)2(2k+3)2-(2k+1)2

(2k+1)2(2k+3)2

=

(2k+3)2-1

(2k+3)2

=

[2(k+1)+1]2-1

[2(k+1)+1]2

由此可知，當(dāng)n=k+1時(shí)等式也成立．根據(jù)（1）（2）可知，等式對(duì)任何n∈N都成立