Question

猜想sn=1+1/(1+2)+1/(1+2+3)+…+1/(1+2+3+…+n)的表达式,并用数学归纳法证明

Answer 1

Tn
=
1
+
2
+
3
+
......
+
n
=
(1/2)n(n
+
1)
1/Tn
=
2/[n(n
+
1)]
=
A/n
+
B/(n
+
1)
=
[(An
+
A)
+
Bn]
=
[A
+
(A
+
B)n]/[n(n
+1)]
由于
A
=
2
企鹅
A
+
B
=
0
所以
B
=
-2
即1/Tn
=
2[1/n
-
1/(n
+
1)]
Sn
=
1/T1
+
1/T2
+
1/T3
+
......
+
1/Tn
=
2[1/1
-
1/(1
+
1)
+
1/2
-
1/(2
+
1)
+
1/3
-
1/(3
+
1)
+
......
+
1/n
-
1/(n
+
1)]
=
2[1
-
1/(n
+
1)]
=
2n/(n
+
1)
1、当n
=
1时，S1
=
2×1/(1
+
1)
=
1，等式成立。
2、设当n
=
k时等式成立，即
Sk
=
2k/(k
+
1)
3、则当n
=
k
+
1时
S(k
+
1)
=
2(k
+
1)/(k
+
1
+
1)
=
2(k
+
1)/[(k
+
1)
+
1]
显然成立，所以有
Sn
=
2n/(n
+
1)

Answer 2

这里用到的裂项相消法
因为1+2+3+..+n=n(n+1)/2
所以[1/（1+2+3+…+n)]=2/n(n+1)=2[1/n-1/(n+1)]
所以Sn=1+[1/(1+2)]+〔1/(1+2+3)〕+[1/(1+2+3+4)]+……+[1/(1+2+3+……+n)]
=2[1/1-1/2]+2[1/2-1/3]+2[1/3-1/4]+...+2[1/n-1/(n+1)]
=2[1-1/2+1/2-1/3+1/3-1/4+...+1/(n-1)-1/n+1/n-1/(n+1)]
=2[1-1/(n+1)]=2n/(n+1)