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a, \(A=2+2^2+2^3+...+2^{90}\)
=> \(A=(2+2^2)+(2^3+2^4)+...+(2^{89}+2^{90})\)
=> \(A=2(1+2)+2^3(1+2)+...+2^{89}(1+2)\)
=> \(A=2.3+2^3.3+...+2^{89}.3\)
=> \(A=(2+2^3+...+2^{89}).3\)chia hết cho 3
b, \(A=2+2^2+2^3+...+2^{90}\)
=> \(A=(2+2^2+2^3)+\left(2^4+2^5+2^6\right)+...+(2^{88}+2^{89}+2^{90})\)
=> \(A=2(1+2+2^2)+2^4.\left(1+2+2^2\right)+...+2^{88}(1+2+2^2)\)
=> \(A=2.7+2^4.7+...+2^{88}.7\)
=> \(A=(2+2^4+...+2^{88}).7\)chia hết cho 7
a, A=2+2^2+2^3+2^4+...+2^90
A=(2+2^2)+(2^3+2^4)+..+(2^89+2^90)
A=2.(1+2)+2^3(1+2)+....+2^89(1+2)
A=2.3+2^3.3+...+2^89.3
A=3.(2+2^3+...+2^89)\(⋮\)3
=> A\(⋮\)3=>ĐPCM
b, A=2+2^2+2^3+....+2^90
A=(2+2^2+2^3)+(2^4+2^5+2^6)+...+(2^88+2^89+2^100)
A=2.(1+2+2^2)+2^4.(1+2+2^2)+...+2^88.(1+2+2^2)
A=2.7+2^4.7+...+2^88.7
A=7.(2+2^4+...+2^88)\(⋮\)7
=>A\(⋮\)7=>ĐPCM
\(B=3+3^2+3^3+...+3^{90}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{89}+3^{90}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{89}\left(1+3\right)\)
\(=\left(1+3\right)\left(3+3^3+...+3^{89}\right)\)
\(=4\left(3+3^3+...+3^{89}\right)⋮4\)
\(B=3+3^2+3^3+...+3^{90}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...\left(3^{88}+3^{89}+3^{90}\right)\)
\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{98}\left(1+3+3^2\right)\)
\(=\left(1+3+3^2\right)\left(3+3^4+...+3^{98}\right)\)
\(=13\left(3+3^4+...+3^{98}\right)⋮13\)
\(A=\frac{2^{12}x3^4x3^{10}}{2^{12}x3^{12}}=3^2=9\)
\(A=\frac{4^6.3^4.9^5}{6^{12}}\)
\(A=\frac{\left(2^2\right)^6.3^4.\left(3^2\right)^5}{\left(2.3\right)^{12}}\)
\(A=\frac{2^{12}.3^4.3^{10}}{2^{12}.3^{12}}\)
\(A=\frac{2^{12}.3^{14}}{2^{12}.3^{12}}\)
\(A=3^2\left(2^{12}.3^{12}\ne0\right)\)
\(A=9\)
Vậy \(A=9\)
Ta có: (n + 5)2 - 3(n + 5) + 2 \(\in\)B(n + 5)
<=> (n + 5)(n + 5 - 3) + 2 \(⋮\)n + 5
<=> 2 \(⋮\)n + 5
<=> n + 5 \(\in\)Ư(2) = {1; -1; 2; -2}
Lập bảng :
| n + 5 | 1 | -1 | 2 | -2 |
| n | -4 | -6 | -3 | -7 |
Vậy ...
Câu c :
2\(^{2x-1}\) - 2 = C
2\(^{2x-1}\)- 2 = 2\(^{101}\)- 2
2\(^{2x-1}\)= 2\(^{101}\)
2x - 1 = 101
2x = 101 + 1 = 102
x = \(\frac{102}{2}\)= 51
Vậy x = 51
a) \(C=2+2^2+2^3+..........+2^{99}+2^{100}\)
\(C=\left(2+2^2+2^3+2^4+2^5\right)+...............+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(C=1.\left(2+2^2+2^3+2^4+2^5\right)+...........+2^{96}.\left(2+2^2+2^3+2^4+2^5\right)\)
\(C=1.62+............+2^{96}.62\)
Mà 62 \(⋮\)31 \(\Rightarrow C⋮31\left(đpcm\right)\)
b) \(2C=2^2+2^3+2^4+2^5+2^6+...............+2^{100}+2^{101}\)
\(2C-C=\left(2^2+2^3+2^4...........+2^{100}+2^{101}\right)-\left(2+2^2+2^3..........2^{99}+2^{100}\right)\)
\(2C-C=2^2+2^3+2^4+...........+2^{100}+2^{101}-2-2^2-2^3-.........-2^{99}-2^{100}\)
\(C=2^{101}-2^{100}\)
c) 22x-1 - 2 = C
Bạn áp dụng phần b để làm
\(B=2+2^2+2^3+...+2^{92}\)
=> \(B=(2+2^2+2^3+2^4)+...+\left(2^{89}+2^{90}+2^{91}+2^{92}\right)\)
=> \(B=2(1+2+2^2+2^3)+...+2^{89}\left(1+2+2^2+2^3\right)\)
=> \(B=2.15+...+2^{89}.15\)
=> \(B=(2+...+2^{89}).15\)CHIA HẾT CHO 15
B = 2 + 22 + 23 + .... + 292
B = ( 2 + 22 + 23 + 24 ) +.....+ ( 289 + 290 + 291 + 292)
B = 2(1+2+22+23) +....+289(1+2+22+23)
B = 2.15+...+289.15
B = (2 + ..... + 2 89)
15 chia hết cho 15