证明Sn=√1x2+√2x3+√3x4+.+√nx(n+1)
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证明Sn=√1x2+√2x3+√3x4+.+√nx(n+1)证明Sn=√1x2+√2x3+√3x4+.+√nx(n+1)证明Sn=√1x2+√2x3+√3x4+.+√nx(n+1)数学归纳法n=1时√
证明Sn=√1x2+√2x3+√3x4+.+√nx(n+1)
证明Sn=√1x2+√2x3+√3x4+.+√nx(n+1)
证明Sn=√1x2+√2x3+√3x4+.+√nx(n+1)
数学归纳法
n=1时
√1x2
用数学归纳法。
(1)当n=1时,S1=√1x2=√2<2,成立
(2)假设n=k时,有Sk=√1x2+√2x3+√3x4+.....+√kx(k+1)<(k+1)²/2
那么 S(k+1)=Sk+√(k+1)(k+2)
<(k+1)²/2+√(k²+3k+2)
<(k+1)²/2+√(k²+3k+9/4)<...
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用数学归纳法。
(1)当n=1时,S1=√1x2=√2<2,成立
(2)假设n=k时,有Sk=√1x2+√2x3+√3x4+.....+√kx(k+1)<(k+1)²/2
那么 S(k+1)=Sk+√(k+1)(k+2)
<(k+1)²/2+√(k²+3k+2)
<(k+1)²/2+√(k²+3k+9/4)
<(k+1)²/2+k+3/2
=(k²+2k+1+2k+3)/2
=(k+2)²/2
从而 n=k+1时,不等式也成立。
所以 Sn=√1x2+√2x3+√3x4+.....+√nx(n+1)<(n+1)²/2 ,n∈N
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证明Sn=√1x2+√2x3+√3x4+.+√nx(n+1)
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