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A = 1 + 3 + 3^2 + ...+ 3 ^100
3A = 3 + 3 ^2 + 3^3 + ... +3^101
3A - A = ( 3 + 3 ^2 + 3^3 + ... +3^101)
- ( 1 + 3 + 3^2 + ...+ 3 ^100)
2A = 3 ^101 - 1
A = \(\frac{\text{ 3 ^101 - 1}}{2}\)
ta có: \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{100^2}=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)
Lại có: \(\frac{1}{2^2}>\frac{1}{2.3};\frac{1}{3^2}>\frac{1}{3.4};\frac{1}{4^2}>\frac{1}{4.5};...;\frac{1}{100^2}>\frac{1}{100.101}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\)
\(=\frac{1}{2}-\frac{1}{101}\)
\(\Rightarrow1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)>1-\left(\frac{1}{2}-\frac{1}{101}\right)=1-\frac{1}{2}+\frac{1}{101}\)
\(=\frac{1}{2}+\frac{1}{101}\)
mà \(\frac{1}{2}=\frac{50}{100}>\frac{1}{100}\Rightarrow\frac{1}{2}+\frac{1}{101}>\frac{1}{100}\)
=> đ p c m
Đặt B= \(2^{99}+2^{98}+...+2^2+2+1\)
\(\Rightarrow2B=2^{100}+2^{99}+2^{98}+...+2^3+2^2+2\)
\(\Rightarrow2B-B=B=2^{100}-1\)
Vậy ta có \(A=2^{100}-B=2^{100}-2^{100}+1=1\)
A = 2100 - ( 299 + 298 + .... + 22 + 2 + 1 )
Đặt B = 299 + 298 + .... + 22 + 2 + 1 , ta có :
B = 299 + 298 + ..... + 22 + 2 + 1
2B = 2100 + 299 + ..... + 23 + 22 + 2
2B - B = ( 2100 + 299 + .... + 23 + 22 + 2 ) - ( 299 + 298 + ..... + 22 + 2 + 1 )
B = 2100 - 1
Vậy A = 2100 - 2100 - 1 = 1
A = 1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + ... + 2018 x 2019
3A = 1 x 2 x ( 3 - 0 ) + 2 x 3 x ( 4 - 1 ) + 3 x 4 x ( 5 -2 ) x ... + 2018 x 2019 x ( 2020 - 2017 )
3A = 1 x 2 x 3 - 1 x 2 x 0 + 2 x 3 x 4 - 2 x 3 x 1 + 3 x 4 x 5 - 3 x 4 x 2 + ... + 2018 x 2019 x 2020 - 2018 x 2019 x 2017
3A = 2018 x 2019 x 2020
A = 2018 x 673 x 2020
A = 2743390280
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)....\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\cdot\cdot\cdot\frac{99}{100}\)
\(=\frac{1.2....99}{2.3....100}=\frac{1}{100}\)
A=(1-1/2)(1-1/3)(1-1/4)....(1-1/100)
A=1/2.2/3.3/4.....99/100
A=(1.2.3....99)/(2.3.4.....100)
A=1/100