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A=100^101+1/100^100+1
B=100^100+1/100^99+1
A<100^101+1+99/100^100+1+99
A<100^101+100/100^100+100
A<100.(100^100+1)/100.(100^99+1)
A<100^100+1/100^99+1=B
=> A<B
Vậy A<B
A= 1-2+3-4+4-5+...+99-100
A = ( 1 - 2 ) + ( 2 - 3 ) + ....+ ( 99 - 100 )
A = ( - 1 ) + ( - 1 ) +....+ ( - 1 )
A = ( - 1 ) . 50
A = - 50
B = 1.2 + 2.3 + 3.4 + 4.5 +...+ 99.100
Nhân cả 2 vế với 3, ta được:
3A=1.2.3+ 2.3.3+ 3.4.3+ 4.5.3+...... 99.100.3
= 1.2.3 + 2.3(4-1) + 3.4.(5-2) +...+ 99.100.(101-98)
= 1.2.3 + 2.3.4 -1.2.3 + 3.4.5-2.3.4 +...+ 99.100.101-98.99.100
= 99.100.101
=) B = (99.100.101) :3
B = 333300
Vậy B= 333300
A= 1-2+3-4+4-5+...+99-100
A = (1-2) + (3-4) + (4-5) + ... + (99-100)
A = (-1) + (-1) + (-1) + ...+ (-1)
A = (-1).50
A = 1
a) \(A=\frac{1}{5}-\frac{1}{5^2}+\frac{1}{5^3}-\frac{1}{5^4}+...+\frac{1}{5^{99}}-\frac{1}{5^{100}}\)
\(\Rightarrow5A=1-\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+...+\frac{1}{5^{98}}-\frac{1}{5^{99}}\)
\(\Rightarrow5A+A=1-\frac{1}{5^{100}}\)
\(A=\frac{1-\frac{1}{5^{100}}}{6}\)
b) B = 1.2+2.3+3.4+...+2017.2018
=>3B=1.2.3 + 2.3.3+3.4.3+...+2017.2018.3
3B = 1.2.3 + 2.3.(4-1) +3.4.(5-2) +...+2017.2018.(2019-2016)
3B = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+2017.2018.2019-2016.2017.2018
3B = 2017.2018.2019
\(B=\frac{2017.2018.2019}{3}\)
3B = 1.2.3 + 2.3.3 + 3.4.3 + ... + 2017.2018.3
3B = 1.2.3 + 2.3.(4-1) + 3.4.(5-2)+...+ 2017.2018(2019-2016)
3B = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 2017.2018.2019 - 2016.2017.2018
3B = 2017.2018.2019
B = 2017.2018.2019/3
B= 2739315938
a, Ta có: \(\frac{-4}{9}=\frac{-8}{18}>\frac{-8}{13}\Rightarrow\frac{-4}{9}>\frac{-8}{13}\)
b,Ta có: \(\frac{-2005}{2006}>-1\)
\(\frac{-2007}{2004}=-1-\frac{3}{2004}<-1\)
\(\Rightarrow-\frac{2005}{2006}>-1>-\frac{2007}{2004}\)
Vậy -2005/2006>-2007/2004