已知一正项数列且Sn=1
S1=a1 =1/2(a1 +1/a1 )
a1²=1
正项数列==>a1=1
S2 =a1+a2 =1+a2 =1/2 (a2 +1/a2)
===>a2=√2 -1
S2 =a1+a2+a3 =√2 +a3
=1/2 (a3 +1/a3)
====>a3 =√3 -√2
猜想,an =√n -√(n-1)
用数学归纳法证明
当n=1时成立已经证明
设n=k时成立,即ak =√k -√(k-1)
则n=k+1时
S(k+1)=Sk+a(k+1) =1/2 [a(k+1) +1/a(k+1)]
1/2 (ak +1/ak)+a(k+1) =1/2 [a(k+1) +1/a(k+1...全部
S1=a1 =1/2(a1 +1/a1 )
a1²=1
正项数列==>a1=1
S2 =a1+a2 =1+a2 =1/2 (a2 +1/a2)
===>a2=√2 -1
S2 =a1+a2+a3 =√2 +a3
=1/2 (a3 +1/a3)
====>a3 =√3 -√2
猜想,an =√n -√(n-1)
用数学归纳法证明
当n=1时成立已经证明
设n=k时成立,即ak =√k -√(k-1)
则n=k+1时
S(k+1)=Sk+a(k+1) =1/2 [a(k+1) +1/a(k+1)]
1/2 (ak +1/ak)+a(k+1) =1/2 [a(k+1) +1/a(k+1)]
ak +1/ak+a(k+1)-1/a(k+1)=0
ak +1/ak =√k -√(k-1) +1/[√k -√(k-1)]
=√k -√(k-1)+(√k +√(k-1)]
=2√k
==>a(k+1)-1/a(k+1)+2√k =0
[a(k+1)]²+2√k a(k+1)-1 =0
解一元而次方程
==>a(k+1)=√(k+1) -√k
或a(k+1)=-√(k+1) -√k (因为是正项数列,舍弃)
所以n=k+1时an =√n -√(n-1)也成立
所以,{an}的通项公式:
an =√n -√(n-1)
答完了才发现,阿炳大师已经先答了,并且方法简单巧妙,很高!
我这个答案打了很久,不撤消了,就当作提供另一个思路吧
。
收起