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học tốt nha
D=12-22+32-42+...+992-1002+1012
D = - (-12 + 22 - 32 + 42 - ... - 992 + 1002) + 1012
D = -[(22 - 12) + (42 - 32) + ... + (1002 - 992)] + 1012
D = -[(2 + 1)(2 - 1) + (4 + 3)(4 - 3) + ... + (100 + 99)(100 - 99)] + 1012
D = -[1 + 2 + 3 + 4 + ... + 99 + 100] + 1012
D = \(-\frac{\left(1+100\right).100}{2}+101^2\)
D = -5050 + 10201
D = 5151
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{1}{1^2}=1\)
\(\frac{1}{2^2}<\frac{1}{1.2}\)
\(\frac{1}{3^2}<\frac{1}{2.3}\)
...
\(\frac{1}{50^2}<\frac{1}{49.50}\)
=> A < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
=> A < \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
=> A < 1 - 1/50 = 49/50
Mà 49/50 < 50/50 = 1 < 2
=> A < 2 (Đpcm).
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S=1+22+23+...+22020
2S= 22+23+24+...+22021
2S - S = S = (22- 22) + (23-23)+ (24- 24)+...+(22020-22020) + (22021-1)
= 22021 - 1
\(S=1+2^2+2^3+...+2^{2020}\)
\(=1+\left(2^2+2^3+...+2^{2020}\right)\). Đặt:
\(A=2^2+2^3+...+2^{2020}\Rightarrow2A=2^3+2^4+...+2^{2021}\)
Do 2A - A = A nên \(A=\left(2^3+2^4+...+2^{2021}\right)-\left(2^2+2^3+...+2^{2020}\right)\)
\(A=2^{2021}-2^2\Rightarrow S=1+\left(2^{2021}-2^2\right)=1+2^{2021}-4\)
Vậy: \(S=1+2^{2021}-4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2, \(=>9A=3^3+3^5+3^7+......+3^{39}+3^{41}\)
\(=>9A-A=3^{41}-3\)
\(=>A=\dfrac{3^{41}-3}{8}\)
CHÚC BẠN HỌC TỐT........
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=> \(3M=3^2+3^3+3^4+...+3^{101}\)
=> \(3M-M=\left(3^2+3^3+3^4+...+3^{101}\right)-\left(3+3^2+3^3+...+3^{100}\right)\)
=> \(2M=3^{101}-3\)
=> \(M=\frac{3^{101}-3}{2}\).
\(2N=2-2^2+2^3-2^4+...-2^{100}+2^{101}\)
=> \(2N-N=\left(2-2^2+2^3-2^4+...-2^{100}+2^{101}\right)-\left(1-2+2^2-2^3+...-2^{99}+2^{100}\right)\)
=> \(N=2^{101}-1\)
M = 3+3^2+3^3+....+3^100
3M = 3^2+3^3+...+3^101
3M - M = (3^2-3^2) + ... + (3^100 - 3^100) + 3^101 - 3
2M = 3^101 - 3
Vậy M = \(\frac{3^{101}-3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=1010+1010^2+1010^3+...+1010^{1011}\)
Suy ra \(1010.S=1010^2+1010^3+1010^4+....+1010^{1012}\)
Nên\(1010.S-S=1010^{1012}-1010\)hay\(1009.S=1010^{1012}-1010\)
Khi đó \(S=\frac{1010^{1012}-1010}{1009}\)
S=1011+1010^2+1010^3+...+1010^1011
S=1+1010+1010^2+1010^3+...+1010^1011
1010.S=1010+1010^2+1010^3+1010^4+...+1010^1012
1010 S - S=1010^1012-1
1009 S=1010^1012-1
S=(1010^1012-1):1009
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Ta có A =1.2 + 2.3 + 3.4 + ...+ 98.99
B = 1^2 + 2^2 + 3^2 +...+98^2 = 1.1+2.2+3.3+...+98.98
Suy ra: A-B= (1.2 + 2.3 + 3.4 + ...+ 98.99) - (1.1+2.2+3.3+...+98.98)
= (1.2-1.1) + (2.3-2.2) + (3.4-3.3) +...+ (98.99-98.98)
= 1(2-1) + 2(3-2) + 3(4-3) +...+ 98(99-98)
= 1.1 + 2.1 + 3.1 +...+ 98.1
= 1+ 2+ 3+...+ 98 = [98.(98+1)]/2= 98.99/2 = 4851
Chúc bạn học tốt!
=> 2A = 22+23+24+...+21011
=> 2A - A = (22+23+24+...+21011) - (21+22+23+...+21010)
=> A = 21011-2.
~ HỌC TỐT NHÁ
A = 2 + 22 + 23 + ... + 22010
2A = 22 + 23 + 24 + ... + 22011
2A - A = 22011 - 2