已知n ∈N,且n>1,求证(1+2/3)(1+2/7)…(1+2/(4n-1))>5/21√(28n+2)
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![已知n ∈N,且n>1,求证(1+2/3)(1+2/7)…(1+2/(4n-1))>5/21√(28n+2)](/uploads/image/z/11557445-5-5.jpg?t=%E5%B7%B2%E7%9F%A5n+%E2%88%88N%2C%E4%B8%94n%3E1%2C%E6%B1%82%E8%AF%81%281%2B2%2F3%29%281%2B2%2F7%29%E2%80%A6%281%2B2%2F%284n-1%29%29%3E5%2F21%E2%88%9A%2828n%2B2%29)
已知n ∈N,且n>1,求证(1+2/3)(1+2/7)…(1+2/(4n-1))>5/21√(28n+2)
已知n ∈N,且n>1,求证(1+2/3)(1+2/7)…(1+2/(4n-1))>5/21√(28n+2)
已知n ∈N,且n>1,求证(1+2/3)(1+2/7)…(1+2/(4n-1))>5/21√(28n+2)
你用数学归纳法做一下,应该是的,可以做出来
当n=1 时 有.成立
假设当n=k时,式子成立,就有
(1+2/3)(1+2/7)…(1+2/(4k+1)>5/21√(28k+1)
在证明能够n=k+1时,有
(1+2/3)(1+2/7)…(1+2/(4(k+1)-1>5/21√(28(k+1)+2)