数列an满足a1=1,an= 根号【 2*a(n-1)的平方 +1】(都在根号内)a1=1,a2=根号3,a3=根号7,猜想an=根号【2^n -1】,写出用数学归纳法推出的具体过程.
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![数列an满足a1=1,an= 根号【 2*a(n-1)的平方 +1】(都在根号内)a1=1,a2=根号3,a3=根号7,猜想an=根号【2^n -1】,写出用数学归纳法推出的具体过程.](/uploads/image/z/7626519-63-9.jpg?t=%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3a1%3D1%2Can%3D+%E6%A0%B9%E5%8F%B7%E3%80%90+2%2Aa%28n-1%29%E7%9A%84%E5%B9%B3%E6%96%B9+%2B1%E3%80%91%EF%BC%88%E9%83%BD%E5%9C%A8%E6%A0%B9%E5%8F%B7%E5%86%85%EF%BC%89a1%3D1%2Ca2%3D%E6%A0%B9%E5%8F%B73%2Ca3%3D%E6%A0%B9%E5%8F%B77%2C%E7%8C%9C%E6%83%B3an%3D%E6%A0%B9%E5%8F%B7%E3%80%902%5En+-1%E3%80%91%2C%E5%86%99%E5%87%BA%E7%94%A8%E6%95%B0%E5%AD%A6%E5%BD%92%E7%BA%B3%E6%B3%95%E6%8E%A8%E5%87%BA%E7%9A%84%E5%85%B7%E4%BD%93%E8%BF%87%E7%A8%8B.)
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数列an满足a1=1,an= 根号【 2*a(n-1)的平方 +1】(都在根号内)a1=1,a2=根号3,a3=根号7,猜想an=根号【2^n -1】,写出用数学归纳法推出的具体过程.
数列an满足a1=1,an= 根号【 2*a(n-1)的平方 +1】(都在根号内)
a1=1,a2=根号3,a3=根号7,猜想an=根号【2^n -1】,写出用数学归纳法推出的具体过程.
数列an满足a1=1,an= 根号【 2*a(n-1)的平方 +1】(都在根号内)a1=1,a2=根号3,a3=根号7,猜想an=根号【2^n -1】,写出用数学归纳法推出的具体过程.
(1)n=1 a1=1 n=2 a2=根号3
(2)设n=k时ak=根号(2^k-1) 成立
(3)那么n=k+1
a(k+1)=根号(2*ak^2 +1)
=根号(2*(2^k-1)+1)
=根号(2^(k+1) -1)
所以得证原式成立.
a1 = 1 =根号(2^1 -1) = 1成立
假设当n=k时成立 当n=k+1时a(k+1)= 根号【 2*a(k)的平方 +1】= 根号(2*(根号(2^k -1))^2 + 1) =根号(2*(2^k -1))+ 1)=2*2^k-2+1 = 2^(k+1) - 1成立所以当n=k+1时假设成立所以结论成立
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