CMR: 1/3+1/3^2+1/3^3+....+1/3^2023< 1/2
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NW
0
11 tháng 4 2023
1+1/2.(1+2)+1/3.(1+2+3)+1/4.(1+2+3+4)+...+1/2023.(1+2+3+...+2023)
=1+1/2.(1+2).2/2+1/3.(1+3).3/2+1/4.(1+4).4/2+...+1/2023.(1+2+3+...+2023).2023/2
=2/2+3/2+4/2+...+2023/2
=2+3+4+...+2023/2
=2025.2022/2/2
=1023637,5
tham khảo thôi nha
VT
9 tháng 4 2023
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1-122+1-132+1-142+....+1-120232"" id="MathJax-Element-2-Frame" role="presentation" tabindex="0" style="box-sizing: inherit; display: inline-block; line-height: 0; font-size: 18.08px; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; position: relative;">A=1−122+1−132+1−1(2+....+1)120232�=1-122+1-132+1-142+....+1-1202321+12+13+...+122023−1
2022-(122+132+142+...+120232)"" id="MathJax-Element-3-Frame" role="presentation" tabindex="0" style="box-sizing: inherit; display: inline-block; line-height: 0; font-size: 18.08px; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; position: relative;">A=2022−(122+132+142+...+120232)�=2022-(122+132+142+...+120232)A
122+132+142+.... <20232
HN
2
AH
Akai Haruma
Giáo viên
15 tháng 4 2023
Lời giải:
Gọi tổng trên là $A$
$A=\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+....+\frac{1}{\frac{2023.2024}{2}}$
$=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2023.2024}$
$=2(\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{2024-2023}{2023.2024})$
$=2(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{2023}-\frac{1}{2024})$
$=2(\frac{1}{3}-\frac{1}{2024})=\frac{2021}{3036}$
NH
0
Set \(S=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2023}}\)
Then \(3S=1+\dfrac{1}{3}+...+\dfrac{1}{3^{2022}}\)
Hence \(2S=3S-S=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{2022}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2023}}\right)\)
\(=1-\dfrac{1}{3^{2023}}\)
\(\Leftrightarrow S=\dfrac{1}{2}-\dfrac{1}{2.3^{2023}}< \dfrac{1}{2}\) (Q. E. D)
Đặt \(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\)
Ta có: \(3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\)
\(3A-A=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\right)\)
\(2A=1-\dfrac{1}{3^{2023}}\)
\(A=\dfrac{1-\dfrac{1}{3^{2023}}}{2}\)
Vì \(\dfrac{1-\dfrac{1}{3^{2023}}}{2}< \dfrac{1}{2}\) nên \(A< \dfrac{1}{2}\)
Vậy...