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2 tháng 11 2023
`#3107.101107`
1.
`a,`
\(A=1+3+3^2+3^3+...+3^{2012}\)
`3A = 3 + 3^2 + 3^3 + ... + 3^2013`
`3A - A = (3 + 3^2 + 3^3 + ... + 3^2013) - (1 + 3 + 3^2 + 3^3 + ... + 3^2012)`
`2A = 3 + 3^2 + 3^3 + ... + 3^2013 - 1 - 3 - 3^2 - 3^3 - ... - 3^2012`
`2A = 3^2013 - 1`
`=> A = (3^2013 - 1)/2`
Vậy, `A = (3^2013 - 1)/2`
`b,`
\(B=1+10+10^2+10^3+...+10^{2023}\)
`10B = 10 + 10^2 + 10^3 + ... + 10^2024`
`10 B - B = (10 + 10^2 + 10^3 + ... + 10^2024) - (1 - 10 + 10^2 + 10^3 + ... + 10^2023)`
`9B = 10 + 10^2 + 10^3 + ... + 10^2024 - 1 - 10^2 - 10^3 - ... - 10^2023`
`9B = 10^2024 - 1`
`=> B = (10^2024 - 1)/9`
Vậy, `B = (10^2024 - 1)/9.`
2 tháng 11 2023
`a)A=1+3+3^2+3^3+...+3^2012`
`=>3A=3+3^2+3^3+...+3^2013`
`=>3A-A=2A=3^2013-1`
`=>A=(3^2013-1)/2`
`b)B=1+10+10^2+...+10^2024`
`=>10B=10+10^2+10^3+....+10^2025`
`=>10B-B=9B=10^2025-10`
`=>B=(10^2025-10)/9`
VN
2
25 tháng 10 2021
a: \(2A=2^1+2^2+...+2^{2022}\)
\(\Leftrightarrow A=2^{2022}-1\)
15 tháng 10 2023
5:
a: \(3^{2n}=\left(3^2\right)^n=9^n\)
\(\left(2^{3n}\right)=\left(2^3\right)^n=8^n\)
=>\(3^{2n}>2^{3n}\)
b: \(199^{20}=\left(199^4\right)^5=1568239201^5\)
\(2003^{15}=\left(2003^3\right)^5=8036054027^5\)
mà \(1568239201< 8036054027\)
nên \(199^{20}< 2003^{15}\)
4: \(100< 5^{2x-1}< 5^6\)
mà \(25< 100< 125\)
nên \(125< 5^{2x-1}< 5^6\)
=>3<2x-1<6
=>4<2x<7
=>2<x<7/2
mà x nguyên
nên x=3
31 tháng 8 2021
\(A=1+2+3+...+7+8=\dfrac{\left(8+1\right).\left(\dfrac{8-1}{1}+1\right)}{2}=36\)
\(B=3+4+5+...+10+11=\dfrac{\left(11+3\right).\left(\dfrac{11-3}{1}+1\right)}{2}=63\)
\(C=1+3+5+...+13+15=\dfrac{\left(15+1\right).\left(\dfrac{15-1}{2}+1\right)}{2}=64\)
\(D=2+4+6+...+18+20=\dfrac{\left(20+2\right).\left(\dfrac{20-2}{2}+1\right)}{2}=110\)
\(E=1+4+7+...+22+25=\dfrac{\left(25+1\right).\left(\dfrac{25-1}{3}+1\right)}{2}=117\)
\(G=1+5+9+...+33+37+41=\dfrac{\left(41+1\right).\left(\dfrac{41-1}{4}+1\right)}{2}=231\)
NH
10 tháng 3 2023
a)
`1/1-1/2`
`=2/2-1/2`
`=1/2`
b)
`1/(1*2)+1/(2*3)`
`=1/1-1/2+1/2-1/3`
`=1/1-1/3`
`=3/3-1/3`
`=2/3`
c)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{1}-\dfrac{1}{100}\\ =\dfrac{99}{100}\)
d)
\(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+...+\dfrac{3}{99\cdot100}\) đề phải như thế này chứ nhỉ?
\(=\dfrac{1\cdot3}{1\cdot2}+\dfrac{1\cdot3}{2\cdot3}+...+\dfrac{1\cdot3}{99\cdot100}\\ =3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =3\cdot\dfrac{99}{100}\\ =\dfrac{297}{100}\)
28 tháng 6 2023
a: 5A=5+5^2+...+5^2023
=>4A=5^2023-1
=>A=(5^2023-1)/4
b: 6B=6^2+6^3+...+6^41
=>5B=6^41-6
=>B=(6^41-6)/5
c: 16C=4^4+4^6+...+4^16
=>15C=4^16-4^2
=>C=(4^16-4^2)/15
d: 9D=3^3+3^5+...+3^27
=>8D=3^27-3
=>D=(3^27-3)/8
4 tháng 4 2022
a: \(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{201}-\dfrac{1}{203}=\dfrac{1}{3}-\dfrac{1}{203}=\dfrac{200}{609}\)
b: \(B=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{73}-\dfrac{1}{76}\)
\(=\dfrac{1}{4}-\dfrac{1}{76}=\dfrac{18}{76}=\dfrac{9}{38}\)
3A=1+1/3+1/3^2+...........+1/3^99
3A-A=1-1/3^100
A=(1-1/3^100):2
A = 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^100
A : 3 = 1/3. ( 1/3 + 1/3^2 + ... + 1/3^100 )
A : 3 = 1/3^2 + 1/3^3 + ... + 1/3^101
A - A : 3 = 1/3 + 1/3^2 + ... + 1/3^100 - 1/3^2 - 1/3^3 - ... - 1/3^101
A . 2/3 = (1/3^2 - 1/3^2) + (1/3^3 - 1/3^3) + ... + (1/3^100 - 1/3^100) + ( 1/3 - 1/3^101 )
A . 2/3 = 0 + 0 + 0 + ... + 0 + 1/3 - 1/3^101
A . 2/3 = 1/3 - 1/3^101
=> A = 1/2 - 1/3^100.2