Question

已知数列1/1×3，1/3×5，1/5×7，...1/(2n－1)(2n＋1)...计算S1S2S3根据计算结果，猜想Sn的表达式，并...

已知数列1/1×3，1/3×5，1/5×7，...1/(2n－1)(2n＋1)...计算S1S2S3根据计算结果，猜想Sn的表达式，并用数学归纳法给出证明

Answer 1

S1=1/(1×3)=1/3
S2=1/(1×3)+1/(3×5)=1/3+1/15=5/15+1/15=6/15=2/5
S3=1/(1×3)+1/(3×5)+1/(5×7)=1/3+1/15+1/35=2/5+1/35=14/35+1/35=15/35=3/7
变形：S1=1/3=1/(2×1 +1) S2=2/5=2/(2×2+1) S3=3/7=(2×3+1)
猜想：Sn=n/(2n+1)
证：
n=1时，S1=1/(2+1)=1/3，与计算结果相等，表达式成立。
假设当n=k(k∈N，且k≥1)时，表达式成立，即Sk=k/(2k+1)，则当n=k+1时，
S(k+1)=1/(1×3)+1/(3×5)+...+1/[(2k-1)(2k+1)]+1/[[2(k+1)-1][2(k+1)+1]]
=Sk+1/[(2k+1)(2k+3)]
=k/(2k+1)+1/[(2k+1)(2k+3)]
=[k(2k+3)+1]/[(2k+1)(2k+3)]
=(2k^2+3k+1)/[(2k+1)(2k+3)]
=(k+1)(2k+1)/[(2k+1)(2k+3)]
=(k+1)/(2k+3)
=(k+1)/[2(k+1)+1]，表达式同样成立。
综上，得Sn=n/(2n +1)。

Answer 2

(1)s1=1/(1*3)=1/3
s2=1/(1*3)+1/(3*5)=(1/2)(1-1/3+1/3-1/5)=2/5
s3=1/(1*3)+1/(3*5)+1/(5*7)=(1/2)(1-1/3+1/3-1/5+1/5-1/7)=3/7
s4=1/(1*3)+1/(3*5)+1/(5*7)+1/(7*9)=(1/2)(1-1/3+1/3-1/5+1/5-1/7+1/7-1/0)=4/9
(2)sn=n/(2n+1)
sn=1/(1*3)+1/(3*5)+1/(5*7)+……+1/[(2n-3)(2n-1)]+1/[(2n-1)(2n+1)]
=(1/2)[1-1/3+1/3-1/5+1/5-1/7+……+1/(2n-3)-1/(2n-1)+1/(2n-1)-1/(2n+1)]
=(1/2)[1-1/(2n+1)]
=(1/2)[2n/(2n+1)
=n/(2n+1)

Answer 3

看看对么