chứng minh 1/3 - 2/3^2 + 3/3^3 + 4/3^4 +.....+99/3^99 - 100/3^100<1/2
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Những câu hỏi liên quan
Chứng minh rằng 1/3-3/2^2+3/3^3-4/3^4+...+99/3^99-100/3^100
chứng minh: 1/3-2/3^2+3/3^3-4/3^4+...+99/3^99-100/3^100<3/16
Chứng minh rằng : 1/3-2/3^2+3/3^3-4/3^4+...+99/3^99-100/3^100<3/16
Xem thêm câu trả lời
- Chứng minh rằng :
1/3 - 2/3^2 + 3/3^3 - 4/3^4 + ... + 99/3^99 - 100/3^100 < 3/16
Cho biểu thức
P= 1/(3)^1 - 2/(3)^2 + 3/(3)^3 - 4/(3)^4 + ... + 99/(3)^99 - 100/(3)^100
Chứng minh rằng P < 3/16
Xem chi tiết
B=1/3+2/3 mũ 2+3/3 mũ 3+4/3 mũ 4+...+99/3 mũ 99+100/ 3 mũ 100 chứng minh B < 3/16
B=1/3+2/3 mũ 2+3/3 mũ 3+4/3 mũ 4+...+99/3 mũ 99+100/ 3 mũ 100 chứng minh B < 3/16
Chứng minh rằng : Z=1/3-2/3^2+3/3^3-4/3^4+.............+99/3^99-100/3^100 + 0+0 <3/16
chứng minh
1/3-2/3^2+3/3^3-4/3^4+.....+99/3^99-100/3^100 <3/16
1 tháng 3 2023 lúc 21:00
Chứng minh: \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)<\(\dfrac{3}{16}\)
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):
\(\Rightarrow16A=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}\)
\(\Rightarrow16A< 3\)
\(\Rightarrow A< \dfrac{3}{16}\)
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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
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