ai bt tự làm

ngu tự chịu

Triệt tiêu hết mấy số kia rồi á bạn

Ta có : 

`5S=5(1/(5^2)+2/(5^3)+3/(5^4)+...+99/(5^100))`

`5S=1/5+2/(5^2)+3/(5^3)+...+99/(5^100)`

`=>5S-S=1/5+2/(5^2)+3/(5^3)+...+99/(5^100)-(1/(5^2)+2/(5^3)+3/(5^4)+...+99/(5^100))`

`4S=1/5+1/(5^2)+1/(5^3)+1/(5^4)+...+1/(5^99) -99/(5^100)`

`20S=5(1/5+1/(5^2)+1/(5^3)+...+1/(5^99)-99/(5^100))`

`20S=1+1/5+1/(5^2)+....+1/(5^98)-99/(5^99)`

`=>20S-4S=(1+1/5+1/(5^2)+...+1/(5^98)-99/(5^99))-(1/5+1/(5^2)+1/(5^3)+...+1/(5^99)-99/(5^100))`

`=>16S=1-99/(5^99)-1/(5^99)-99/(5^100)`

Vì `-99/(5^99)-1/(5^99)-99/(5^100)<0=>1-99/(5^99)-1/(5^99)-99/(5^100)<1`

`=>S<1/16`

minh chiu kho qua thong cam nha !!!!!!!!!!!!!! hi hi

Ta có: \(S=\frac{1}{5^2}+\frac{2}{5^3}+\cdots+\frac{99}{5^{100}}\)

=>\(5S=\frac15+\frac{2}{5^2}+\cdots+\frac{99}{5^{99}}\)

=>\(5S-S=\frac15+\frac{2}{5^2}+\cdots+\frac{99}{5^{99}}-\frac{1}{5^2}-\frac{2}{5^3}-\cdots-\frac{99}{5^{100}}=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

=>\(4S=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

Ta có: \(A=\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}\)

=>\(5A=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{98}}\)

=>\(5A-A=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{98}}-\frac{1}{5^2}-\frac{1}{5^3}-\cdots-\frac{1}{5^{99}}\)

=>\(4A=\frac15-\frac{1}{5^{99}}=\frac{5^{98}-1}{5^{99}}\)

=>\(A=\frac{5^{98}-1}{4\cdot5^{99}}\)

Ta có: \(4S=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

\(=\frac15+\frac{5^{98}-1}{4\cdot5^{99}}-\frac{99}{5^{100}}=\frac15+\frac{5^{99}-5-396}{4\cdot5^{100}}=\frac15+\frac{1}{4\cdot5}-\frac{401}{4\cdot5^{100}}\)

=>\(4S<\frac15+\frac{1}{20}=\frac{4}{20}+\frac{1}{20}=\frac{5}{20}=\frac14\)

hay S<1/16

Ta có: \(S=\frac{1}{5^2}+\frac{2}{5^3}+\cdots+\frac{99}{5^{100}}\)

=>\(5S=\frac15+\frac{2}{5^2}+\cdots+\frac{99}{5^{99}}\)

=>\(5S-S=\frac15+\frac{2}{5^2}+\cdots+\frac{99}{5^{99}}-\frac{1}{5^2}-\frac{2}{5^3}-\cdots-\frac{99}{5^{100}}=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

=>\(4S=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

Ta có: \(A=\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}\)

=>\(5A=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{98}}\)

=>\(5A-A=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{98}}-\frac{1}{5^2}-\frac{1}{5^3}-\cdots-\frac{1}{5^{99}}\)

=>\(4A=\frac15-\frac{1}{5^{99}}=\frac{5^{98}-1}{5^{99}}\)

=>\(A=\frac{5^{98}-1}{4\cdot5^{99}}\)

Ta có: \(4S=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

\(=\frac15+\frac{5^{98}-1}{4\cdot5^{99}}-\frac{99}{5^{100}}=\frac15+\frac{5^{99}-5-396}{4\cdot5^{100}}=\frac15+\frac{1}{4\cdot5}-\frac{401}{4\cdot5^{100}}\)

=>\(4S<\frac15+\frac{1}{20}=\frac{4}{20}+\frac{1}{20}=\frac{5}{20}=\frac14\)

hay S<1/16

a, S= 1/1*2 + 1/2*3 + 1/3*4 +...+1/99*100
    S= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/99 - 1/100
    S= 1/1 - 1/100
    S= 100/100 - 1/100
    S= 99/100

b, S= 1/1*3 + 1/3*5 + 1/5*7 +...+1/99*101
    S= 1/2* (2/1*3 + 2/3*5 + 2/5*7 +...+ 2/99*101)
    S= 1/2* (1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 +...+ 1/99 - 1/101)
    S= 1/2* (1/1 - 1/101)
    S= 1/2* (101/101 - 1/101)
    S= 1/2* 100/101
    S= 50/101
Chúc bạn học tốt nha

\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)

   \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)

     \(=1-\frac{1}{100}=\frac{99}{100}\)