问题
解答题
已知数列an=1+
|
答案
证明:当n=1时,a1=1
S1=a1=1满足条件
假设当n=k,(k>1,k∈N)时Sk=(k+1)ak-k成立
当n=k+1时,
∵ak=1+
+1 2
+…+1 3
=1+1 k
+1 2
+…+1 3
+1 k
-1 k+1
=ak+1-1 k+1 1 k+1
则Sk+1=Sk+ak+1=(k+1)ak-k+ak+1=(k+1)(ak+1-
)-k+ak+11 k+1
=(k+1)ak+1-1-k+ak+1=(k+2)ak+1-(1+k)
从而Sn=(n+1)an-n成立.
得证.