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Ta có: 5M - M = 5^2001 - 1
4M = 5^2001 - 1
(4M+1) = 5^2001
Ta có : 5^2001 * 2^2010
5^2001 = .....25 ( số tự nhiên)
2^2010 = (2^20)^100 * 2^10
= 76^100 * 1024
= ....76( số tự nhiên) * 1024
= ......24
Vay 5^2001 * 2^2010 = ....25 * ....24
= .....00 chia het cho 2 va 4
Vậy số trên là số chính phương.
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TA CÓ:
A=30+3+32+33+........+311
(30+3+32+33)+....+(38+39+310+311)
3(0+1+3+32)+......+38(0+1+3+32)
3.13+....+38.13 cHIA HẾT CHO 13 NÊN A CHIA HẾT CHO 13( đpcm)
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Ta có:
Xét số a. Ta có a2 > (a - 1)(a + 1)
Thật vậy, (a - 1)(a + 1) = a(a + 1) - (a + 1) = a2 + a - a - 1 = a2 - 1 < a2
Suy ra \(\dfrac{1}{\left(a-1\right)\left(a+1\right)}>\dfrac{1}{a^2}\)
Ta có:
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
\(< \dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{99.101}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{100}-\dfrac{1}{101}\right)\)
\(< \dfrac{3}{4}\)
Ko bt có sai chỗ nào ko....
![](https://rs.olm.vn/images/avt/0.png?1311)
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Có:1/2^2+1/3^2+...+1/20^2<1/1*2+1/2*3+...+1/19*20=1-1/20=19/20<1
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Co 1/2^2+1/3^2+...+1/100^2<1/1.2+1/2.3+...+1/99.100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100<1
vay 1/2^2+...+1/100^2<1
0,75 = \(\dfrac{3}{4}\)
Ta có: \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{2000^2}\) < \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ... +\(\dfrac{1}{2000.2001}\).
<=> \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{2000^2}\) < \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{2000}\) - \(\dfrac{1}{2001}\).
<=> \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{2000^2}\) < \(\dfrac{1}{2}\) - \(\dfrac{1}{2001}\).
Vì \(\dfrac{1}{2}\) < \(\dfrac{3}{4}\) nên \(\dfrac{1}{2}\) - \(\dfrac{1}{2001}\) < \(\dfrac{3}{4}\).
Vậy \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{2000^2}\) < \(\dfrac{3}{4}\).