已知sn=3^n-2求an

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 。
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