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1.
a.Để A là phân số thì n - 5 khác 0 => n khác 5
b.Để A \(\in\)Z thì 3 chia hết cho n - 5 => n - 5 \(\in\) Ư(3) = {1; 3; -1; -3}
Ta có bảng sau:
| n - 5 | 1 | -1 | 3 | -3 |
| n | 6 | 4 | 8 | 2 |
Vậy n \(\in\){6; 4; 8; 2} thì A \(\in\)Z.
a)A=\(\frac{\left(8+100\right).\left[\left(100-8\right):4+1\right]}{2}=\frac{108.242}{2}=13068\)
b) \(5B=5^2+5^3+...+5^{101}\)
\(5B-B=5^{101}-5\)
\(B=\frac{5^{101}-5}{4}\)
ta có: \(A=1+4+4^2+4^3+...+4^{99}\)
\(\Leftrightarrow4A=1.4+4.4+4^2.4+4^3.4+...+4^{99}.4\)
\(\Leftrightarrow4A=4+4^2+4^3+4^4+...+4^{100}\)
\(\Leftrightarrow4A-A=\left(4+4^2+4^3+4^4+...+4^{100}\right)-\left(1+4+4^2+4^3+...+4^{99}\right)\)
\(\Leftrightarrow3A=4^{100}-1\)
\(\Leftrightarrow3A=B-1\)
\(\Leftrightarrow A=\frac{B-1}{3}\)
Mà:\(\frac{B-1}{3}< \frac{B}{3}\)
Nên:\(A< \frac{B}{3}\)
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a) Ta có: \(B=1+3+3^2+....+3^{2006}\)
\(\Leftrightarrow3B=3+3^2+.....+3^{2006}+3^{2007}\)
\(\Rightarrow3B-B=3^{2007}-1\)
\(\Leftrightarrow B=\dfrac{3^{2007}-1}{2}\)
Vậy \(B=\dfrac{2^{2007}-1}{2}\)
b) Ta có: \(A=3^{2007}-1=\left(3-1\right)\left(3^{2006}+3^{2005}+.......+3+1\right)\)
\(\Leftrightarrow A=2\left(3^{2006}+3^{2005}+....+3+1\right)\) luôn chia hết cho 2
Vậy \(A=\left(3^{2007}-1\right)⋮2\)
a) \(B=1+3+3^2+3^3+3^4+.......+3^{2006}\)
\(\Leftrightarrow3B=3+3^2+3^3+3^4+.......+3^{2007}\)
\(\Leftrightarrow3B-B=\left(3+3^2+3^3+3^4+.......+3^{2007}\right)-\left(1+3+3^2+3^3+3^4+.......+3^{2006}\right)\)
\(\Leftrightarrow2B=3^{2007}-1\)
\(\Leftrightarrow B=\dfrac{3^{2007}-1}{2}\)
Vậy \(B=\dfrac{3^{2007}-1}{2}\)
Mời bạn tham khảo các link sau:
a),b),c):https://hoidap247.com/cau-hoi/214111
d):https://olm.vn/hoi-dap/detail/78449788871.html
B1 : B-A = 1/2
B2 :
CM được : A = (4^100-1)/3
=> A < 4^100/3 = B/3
Tk mk nha
Bài 1 :
A = 1 + 3 + 32 + 33 + ....... + 320
\(\Rightarrow3A=3+3^2+3^3+3^4+......+3^{21}\)
\(\Rightarrow3A-A=\left(3+3^2+3^3+3^4+.....+3^{21}\right)-\left(1+3+3^2+3^3+......+3^{20}\right)\)
\(\Rightarrow2A=2+3^{21}\)
\(\Rightarrow A=\frac{2+3^{21}}{2}\)
\(\Rightarrow B-A=\left(2+3^{21}\right):2-3^{21}:2\)
\(\Rightarrow B-A=1+3^{21}:2-3^{21}:2\)
\(\Rightarrow B-A=1+\left(3^{21}:2-3^{21}:2\right)\)
\(\Rightarrow B-A=1+0\)
\(\Rightarrow B-A=1\)
Vậy \(B-A=1\)
Bài 2 :
\(A=1+4+4^2+4^3+.....+4^{99}\)
\(\Rightarrow4A=4+4^2+4^3+4^4+.....+4^{100}\)
\(\Rightarrow4A-A=\left(4+4^2+4^3+4^4+.....+4^{100}\right)-\left(1+4+4^2+4^3+......+4^{99}\right)\)
\(\Rightarrow3A=3+4^{100}\)
\(\Rightarrow A=\frac{3+4^{100}}{3}\)
\(\Rightarrow\frac{B}{3}=\frac{4^{100}}{3}\)
Vì \(4^{100}=4^{100}\)nên \(3+4^{100}>4^{100}\)
Vậy \(A>\frac{B}{3}\left(ĐPCM\right)\)
1. Tính tổng:
B = 2 - 4 - 6 + 8 + 10 - 12 - 14 + 16 + ... + 2002 - 2004 - 2006 + 2008
=> ( 2 - 4 - 6 + 8 )+ (10 - 12 - 14 + 16) + ... + (2002 - 2004 - 2006 + 2008)
=> (-8+ 8) +(-16+ 16) +.........+ ( -2008+ 2008)(1)
=> 0+0+...........+0
=> 0
Ta thấy (1) đều là những số đối nên kết quả đường nhiên bằng 0
\(A=1+4+4^2+4^3+...+4^{99}\\ \Rightarrow4A=4+4^2+4^3+...+4^{100}\\ \Rightarrow3.A=4^{100}-1\\ \Rightarrow A=\dfrac{4^{100}-1}{3}< \dfrac{4^{100}}{3}=\dfrac{B}{3}\\ \Rightarrow A< \dfrac{B}{3}\)