求证:an=[√(4n-3)-1]/21/(a1+1)+1/(a1+1)(a2+1)+1/(a1+1)(a2+1)(a3+1)+.+1/(a1+1)(a2+1)(a3+1)...(an+1)<3对于任意n包含于N+,都有an=[√(4n-3)-1]/2求证1/(a1+1)+1/(a1+1)(a2+1)+1/(a1+1)(a2+1)(a3+1)+......+1/(a1+1)(a2+1)(a3+1)...(an

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求证:an=[√(4n-3)-1]/21/(a1+1)+1/(a1+1)(a2+1)+1/(a1+1)(a2+1)(a3+1)+.+1/(a1+1)(a2+1)(a3+1)...(an+1)<3对于任意n包含于N+,都有an=[√(4n-3)-1]/2求证1/(a1+1)+1/(a1+1)(a2+1)+1/(a1+1)(a2+1)(a3+1)+......+1/(a1+1)(a2+1)(a3+1)...(an
求证:an=[√(4n-3)-1]/2
1/(a1+1)+1/(a1+1)(a2+1)+1/(a1+1)(a2+1)(a3+1)+.+1/(a1+1)(a2+1)(a3+1)...(an+1)<3
对于任意n包含于N+,都有an=[√(4n-3)-1]/2
求证1/(a1+1)+1/(a1+1)(a2+1)+1/(a1+1)(a2+1)(a3+1)+......+1/(a1+1)(a2+1)(a3+1)...(an+1)<3

求证:an=[√(4n-3)-1]/21/(a1+1)+1/(a1+1)(a2+1)+1/(a1+1)(a2+1)(a3+1)+.+1/(a1+1)(a2+1)(a3+1)...(an+1)<3对于任意n包含于N+,都有an=[√(4n-3)-1]/2求证1/(a1+1)+1/(a1+1)(a2+1)+1/(a1+1)(a2+1)(a3+1)+......+1/(a1+1)(a2+1)(a3+1)...(an
1/(an+1)=2/((4n-3)^(1/2)+1)<2/(2+1)=2/3
下来直接代进去,成2/3等比数列的求和

用数学归纳法

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