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b) ...........
= 1/2x2/3x.....x999/1000
= 1x2x...x999/2x3x...x1000
=1/1000
a) ...............
= 3/2 . 4/3 .... 1000/999
= 3x4x5x....x1000/2x3x4x...x999
=1000/2=500
a, \(\left(\dfrac{1}{2}+1\right).\left(\dfrac{1}{3}+1\right).\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{999}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{1000}{999}\)
\(=\dfrac{3.4.5...1000}{2.3.4...999}\)
\(=\dfrac{1000}{2}\)\(=500\)
b, \(\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{1000}-1\right)\)
\(=\dfrac{-1}{2}.\dfrac{-2}{3}.\dfrac{-3}{4}...\dfrac{-999}{1000}\)
\(=\dfrac{\left(-1\right).\left(-2\right).\left(-3\right)...\left(-999\right)}{2.3.4...1000}\)
\(=\dfrac{-1}{1000}\)
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3A = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{39}}\)
A = \(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{40}}\)
=> 2A = 3A - A = \(1-\frac{1}{3^{40}}\)=> \(\frac{1-\frac{1}{3^{40}}}{2}=\frac{1}{2}-\frac{1}{3^{40}\cdot2}\)
Mấy câu còn là thì tương tự nhé c
câu b nhân vào \(2^2\)
câu c nhân vào 4
A = 2^3 + 2^4+ 2^5+ 2^6 + 2^7 + ... + 2^90
2A = 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + .... + 2^90 + 2^100
2A - A = ( 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + .... + 2^90 + 2^100 ) - ( 2^3 + 2^4+ 2^5+ 2^6 + 2^7 + ... + 2^90 )
A = 2^100 - 2^3
B = 1 + 5 + 5^2 + 5^3 + 5^4 + .... + 5^50
5B = 5 + 5^2 + 5^3 + 5^4 + 5^5 + .... + 5^50 + 5^51
5B - B = ( 5 + 5^2 + 5^3 + 5^4 + 5^5 + .... + 5^50 + 5^51 ) - ( 1 + 5 + 5^2 + 5^3 + 5^4 + .... + 5^50 )
4B = 5^51 - 1
B = 5^51 - 1 / 4
a,\(\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{999}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}.....\dfrac{1000}{999}\)
\(=\dfrac{3.4.5....1000}{2.3.4....999}=\dfrac{1000}{2}=500\)
b,\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{1000}-1\right)\)
\(=\dfrac{-1}{2}.\dfrac{-2}{3}.\dfrac{-3}{4}.....\dfrac{-999}{1000}\)
=\(\dfrac{-\left(1.2.3....999\right)}{2.3.4....1000}=\dfrac{-1}{1000}\)
c,\(\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}....\dfrac{99}{10^2}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}....\dfrac{9.11}{10.10}\)
\(=\dfrac{1.3.2.4.3.5....9.11}{2.2.3.3.4.4....10.10}\)
\(=\dfrac{1.2.3...9}{2.3.4...10}.\dfrac{3.4.5...11}{2.3.4...10}\)
\(=\dfrac{1}{10}.\dfrac{11}{2}=\dfrac{11}{20}\)