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1,2 dễ ko làm
3,
S = 1 + 2 + 22 + 23 + ... + 29
2S = 2 + 22 + 23 + 24 + ... + 210
2S - S = ( 2 + 22 + 23 + 24 + ... + 210 ) - ( 1 + 2 + 22 + 23 + ... + 29 )
S = 210 - 1
Mà 5 . 28 = ( 1 + 22 ) . 28 = 28 + 210 > 210 > 210 - 1
Vậy S < 5 . 28
P = 1 + 3 + 32 + 33 + ... + 320
3P = 3 + 32 + 33 + 34 + ... + 321
3P - P = ( 3 + 32 + 33 + 34 + ... + 321 ) - ( 1 + 3 + 32 + 33 + ... + 320 )
2P = 321 - 1
P = ( 321 - 1 ) : 2 < 321
Vậy P < 321
Ta có A = 1 + 2 + 22 + 23 + ... + 2100
=> 2A = 2 + 22 + 23 + 24 + ... + 2101
Khi đó 2A - A = (2 + 22 + 23 + 24 + ... + 2101) - (1 + 2 + 22 + 23 + ... + 2100)
=> A = 2101 - 1
Vì 2101 - 1 < 2101
=> A < B
Vậy A < B
A = 1 + 2 + 22 + 23 + ... + 2100
=> 2A = 2( 1 + 2 + 22 + 23 + ... + 2100 )
= 2 + 22 + 23 + ... + 2101
=> A = 2A - A
= 2 + 22 + 23 + ... + 2101 - ( 1 + 2 + 22 + 23 + ... + 2100 )
= 2 + 22 + 23 + ... + 2101 - 1 - 2 - 22 - 23 - ... - 2100
= 2101 - 1 < 2101
=> A < B
\(B=\frac{2^{2020}+2}{2^{2021}+2}=\frac{2\left(2^{2019}+1\right)}{2\left(2^{2020}+1\right)}=\frac{2^{2019}+1}{2^{2020}+1}\)
vậy A=B=\(\frac{2^{2019}+1}{2^{2020}+1}\)
\(B=\frac{2^{2020}+2}{2^{2021}+2}\)
\(=\frac{2\left(2^{2019}+1\right)}{2\left(2^{2020}+1\right)}\)
\(=\frac{2^{2019}+1}{2^{2020}+1}=A\)
Vậy \(A=B\)
P/s: Bài này mk thường thấy dạng như phía dưới, bn đọc tham khảo
\(B=\frac{2^{2020}+1}{2^{2021}+1}< \frac{2^{2020}+1+1}{2^{2021}+1+1}=\frac{2^{2020}+2}{2^{2021}+2}=\frac{2^{2019}+1}{2^{2020}+1}=A\)
Vậy \(A>B\)
\(3,1+5^2+5^4+...+5^{26}\)
\(=\left(1+5^2\right)+\left(5^4+5^6\right)+...+\left(5^{24}+5^{26}\right)\)
\(=\left(1+5^2\right)+5^4\left(1+5^2\right)+...+5^{24}\left(1+5^2\right)\)
\(=26+5^4.26+...+5^{24}.26\)
\(=26\left(5^4+...+5^{24}\right)\)
Vì \(26⋮26\)
\(\Rightarrow26\left(5^4+...+5^{24}\right)⋮26\)
\(\Rightarrow1+5^2+5^4+...+5^{26}⋮26\)
\(4,1+2^2+2^4+...+2^{100}\)
\(=\left(1+2^2+2^4\right)+...+\left(2^{98}+2^{99}+2^{100}\right)\)
\(=\left(1+2^2+2^4\right)+....+2^{98}\left(1+2^2+2^4\right)\)
\(=21+2^6.21...+2^{98}.21\)
\(=21\left(2^6+...+2^{98}\right)\)
Có : \(21\left(2^6+...+2^{98}\right)⋮21\)
\(\Rightarrow1+2^2+2^4+...+2^{100}⋮21\)
\(2^2>1.3\); \(3^2>2.4\) ; \(n^2>\left(n-1\right)\left(n+1\right)\)
\(\Rightarrow A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{2018.2020}\)
\(A< \frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2018}-\frac{1}{2020}\right)\)
\(A< \frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{2020}\right)< \frac{1}{2}\left(1+\frac{1}{2}\right)=\frac{3}{4}\)
\(\Rightarrow A< \frac{3}{4}\)