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`#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.`
`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`
(1981 x 1982 - 990) : (1980 x 1982 + 992)
=(1980 x 1982+1982 -990) : (1980 x 1982 +992)
=(1980 x 1982 + 992) : ( 1980 x 1982 + 992)
=1
B=[(45.79+45.21)]:90-5^2]:5+2^3 B=[(45.79+45.21):90-25]:5+8 B=[(45.(79+21):65]:13 B=[(45.100):65]:13 B=[4500:65]:13 B=4500:65:13
a, Ta có : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
=> \(\frac{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}=1\)
=> đpcm
Study well ! >_<
\(A=2^0+2^1+2^2+...+2^{2010}\)
\(2A=2^1+2^2+2^3+...+2^{2021}\)
\(2A-A=\left(2^1+2^2+2^3+...+2^{2021}\right)-\left(2^0+2^1+2^2+...+2^{2020}\right)\)
\(A=2^{2021}-1\)
Câu 1
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{99}=\frac{49}{100}\)
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