问题
填空题
设f(n)=1+
|
答案
∵f(n)=1+
+1 2
+…+1 3
,1 n
∴f(2k+1)-f(2k)=1+
+1 2
+…+ 1 3
+1 2k
+1 2k+1
+…+1 2k+2
-(1+1 2k+ 2k
+1 2
+…+1 3
)1 2k
=
+1 2k+1
+…+1 2k+2
.1 2k+1
故答案为:
+1 2k+1
+…+1 2k+2
.1 2k+1