Ta có: \(S=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

=>\(3S=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\ldots+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

=>3S+S=\(1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{99}{3^{98}}-\frac{100}{3^{99}}+\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

=>4S=\(1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

Đặt \(A=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}\)

=>3A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}\)

=>3A+A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}\)

=>4A=\(-1-\frac{1}{3^{99}}=\frac{-3^{99}-1}{3^{99}}\)

=>\(A=\frac{-3^{99}-1}{4\cdot3^{99}}\)

Ta có: \(4S=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(=1+\frac{-3^{99}-1}{4\cdot3^{99}}-\frac{100}{3^{100}}=1+\frac{-3^{100}-3-400}{4\cdot3^{100}}=1-\frac14-\frac{403}{4\cdot3^{100}}<\frac34\)

=>\(S<\frac{3}{16}\)

mà 3/16<3/15=1/5

nên S<1/5

2²⁵⁰ > 3¹⁰⁰

Ta có :

`2^250 = ( 2^2 )^{125} = 4^{125}`

Do `3^{100} < 4^{100}<4^{125} => 3^{100}<4^{125}=>2^{250}>3^{100}`

Vậy `2^{250}>3^{100}`

Đề sai nhé: phải là 8S-..+1 nhé

Có: \(3^2.S=3^2+3^4+3^6+...+3^{2004}\)

\(\Rightarrow3^2S-S=3^{2004}-1\)\(\Leftrightarrow8S=3^{2004}-1\Leftrightarrow8S-3^{2004}+1=0\)

S=4+4²+4³+4⁴+...+4⁹⁹

4S=4²+4³+4⁴+...+4⁹⁹+4¹⁰⁰

4S-S=4¹⁰⁰-1

3S=4¹⁰⁰-1

S=(4¹⁰⁰-1):3 < 6.4⁹⁸

vậy S < 6.4⁹⁸