Đúng rồi đó ngu còn bày đặt

Lời giải:

\(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

\(3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

\(A+3A=1+\frac{1-2}{3}+\frac{-2+3}{3^2}+\frac{3-4}{3^3}+\frac{-4+5}{3^4}+...+\frac{99-100}{3^{99}}-\frac{100}{3^{100}}\)

\(4A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-.....+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(4A=(1-\frac{1}{3})+(\frac{1}{3^2}-\frac{1}{3^3})+...+(\frac{1}{3^{98}}-\frac{1}{3^{99}})-\frac{100}{3^{100}}\)

\(4A=\frac{2}{3}+\frac{2}{3^3}+...+\frac{2}{3^{99}}-\frac{100}{3^{100}}\)

\(2A=\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{50}{3^{100}}\)

\(18A=3+\frac{1}{3}+...+\frac{1}{3^{97}}-\frac{450}{3^{100}}\)

\(\Rightarrow 18A-2A=3-\frac{1}{3^{99}}-\frac{450}{3^{100}}+\frac{50}{3^{100}}=3-\frac{1}{3^{99}}-\frac{400}{3^{100}}\)

\(\Leftrightarrow 16A=3-\frac{1}{3^{99}}-\frac{400}{3^{100}}<3\Rightarrow A< \frac{3}{16}\)

Đặt A=1/3-2/3^2+3/3^3-4/3^4+...+99/3^99-100/3^100

3A=1-2/3+3/3^2-4/3^3+...+99/3^98-100/3^99

3A+A=1-1/3+1/3^2-1/3^3+1/3^4-...+1/3^98-1/3^99-100/3^100

<1-1/3+1/3^2-1/3^3+1/3^4-...+1/3^98-1/3^99

Đặt S=1-1/3+1/3^2-1/3^3+1/3^4-...+1/3^98-1/3^99

3S=3-1+1/3-1/3^2+1/3^3-...-1/3^98

3S+S=3-1/3^99

S=(3-1/3^99) :4

S=3/4-1/4.3^99

\(\Rightarrow\)4A<3/4-1/4.3^99

\(\Rightarrow\)A<(3/4-1/4.3^99):4

\(\Rightarrow\)A<3/16-1/16.3^99<3/16

Vậy 1/3-2/3^2+3/3^3-4/3^4+...+99/3^99-100/3^100<3/16

Đặt \(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)

\(\Rightarrow3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)

\(\Rightarrow A+3A=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)

\(\Rightarrow4A=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\) (1)

\(\Rightarrow12A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{97}}-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\) (2)

Cộng vế (1) và (2):

\(\Rightarrow16A=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}\)

\(\Rightarrow16A< 3\)

\(\Rightarrow A< \dfrac{3}{16}\)

Đặt `A` `=` `1/3 - 2/3^2+3/3^3 - 4/3^4+ ... + 99/3^99-100/3^100`
`=>3A=1 -2/3 +3/3^2 - 4/3^3+ ... - 100/3^99`
`=>4A=A+3A=1-1/3+1/3^2-1/3^3+...-1/3^99 - 100/3^100`
`=>12A=3.4A=3-1+1/3-1/3^2+...-1/3^98 - 100/3^99`
`=>16A=12A+4A=3-1/3^99-100/3^99-100/3^1...`
`=>16A=3-101/3^99-100/3^100`
`<=>A=3/16-(101/3^99+100/3^100)/16 < 3/16`
`=> A<3/16`

@Nae

Đáp án nè:

Đặt A=\(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{99}}\)

3A=\(\dfrac{1}{1}-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)

3A+A=\(\left(\dfrac{1}{1}-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\right)\)

4A=\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}-\dfrac{1}{3^{100}}\)

4A bé hơn(sorry tớ không thấy dấu bé hơn)\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)

Đặt B=\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)

3B=\(3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\)

4B=\(3-\dfrac{1}{3^{99}}\) bé hơn 3 \(\Rightarrow\)B bé hơn \(\dfrac{3}{4}\)

\(\Rightarrow\) 4A bé hơn\(\dfrac{3}{4}\Rightarrow\)A bé hơn \(\dfrac{3}{16}\)

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