CM:(n!)^2>n^n
K
Khách
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TA
31 tháng 8 2020
Xét số nguyên dương thỏa mãn điều kiện \(1\le k< n-1\)
\(\Leftrightarrow n-k-1>0\Leftrightarrow nk-k^2-k>0\Leftrightarrow nk-k^2+n-k-n>0\)
\(\Leftrightarrow k\left(n-k\right)+n-k>n\Leftrightarrow\left(k+1\right)\left(n-k\right)>n\)
Lần lượt cho k = 1, 2, 3, ..., ( n - 2 ):
Với n > 2, ta có: \(2\left(n-1\right)>n\)
\(3\left(n-2\right)>n\)
\(4\left(n-3\right)>n\)
\(................\)
\(\left(n-1\right)\left[n-\left(n-2\right)\right]>n\)
\(\Leftrightarrow2.3.4...\left(n-1\right).2.3.4...\left(n-1\right)>n^{n-2}\)
\(\Leftrightarrow\left[2.3.4...\left(n-1\right)\right]^2>n^{n-2}\)
\(\Leftrightarrow\left[\left(n-1\right)!\right]^2>n^{n-2}\)
Nhân 2 vế với \(n^2\), ta có: \(\left(n!\right)^2>n^2\left(đpcm\right)\)
DA
31 tháng 8 2020
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HB
0
NB
0
4 tháng 1 2020
Em học lớp 8 thôi :)) Cái này em k chắc lắm ạ, có gì sai anh chỉ nhé !
Gợi ý :
3) \(n^3+11n=n\cdot\left(n^2+11\right)=n\cdot\left(n^2-1+12\right)\)
\(=n\left(n-1\right)\left(n+1\right)+12n⋮6\)
1) \(Có:2^n-2n-1=2\left(2^{n-1}-1\right)-1>0\forall n\ge3\)
nên : \(2^n>2n+1\)
TN
0
TN
1
DD
Đoàn Đức Hà
Giáo viên
6 tháng 2 2021
Ta có: \(n^6-n^4+2n^3+2n^2=n^2\left(n^4-n^2+2n+2\right)=n^2\left[n^2\left(n-1\right)\left(n+1\right)+2\left(n+1\right)\right]\)
\(=n^2\left(n+1\right)\left(n^3-n^2+2\right)=n^2\left(n+1\right)\left(n^3+n^2-2n^2+2\right)=n^2\left(n+1\right)\left[n^2\left(n+1\right)-2\left(n+1\right)\left(n-1\right)\right]\)\(=n^2\left(n+1\right)^2\left(n^2-2n+2\right)\)
Để \(A\)là số chính phương thì \(n^2-2n+2\)là số chính phương.
Ta có: \(n^2-2n+2< n^2\)(do \(n>1\))
\(n^2-2n+2=\left(n-1\right)^2+1>\left(n-1\right)^2\)
\(\Rightarrow\left(n-1\right)^2< n^2-2n+2< n^2\)nên \(n^2-2n+2\)không thể là số chính phương.
Vậy \(A=n^6-n^4+2n^3+2n^2\)không là số chính phương.
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