cho A=1/2 mũ 2+1/2 mũ 3+1/2 mũ 4+...+1/2 mũ 2020+1/2 mũ 2021. chúng tỏ A<1/2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
27 tháng 8 2023
Bài 1 :
\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)
\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)
\(\Rightarrow M< N\)
27 tháng 8 2023
Bài 3 :
a) \(t^2+5t-8\) khi \(t=2\)
\(=5^2+2.5-8\)
\(=25+10-8\)
\(=27\)
b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)
\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)
\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)
c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)
\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)
\(\left(1\right)=1^3=1\)
HN
1
VT
29 tháng 11 2021
A=(1+3+32)+(33+34+35)+...+(32019+32020+32021) A=(1+3+32)+33.(1+3+32)+...+32019.(1+3+32)
A=13+33.13+...+32019.13
A=13.(1+33+...+32019)chia hết cho 13
=>A chia hết cho 13
DD
Đoàn Đức Hà
Giáo viên
21 tháng 5 2021
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{10-9}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}< 1\)
HN
21 tháng 5 2021
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\\ A< \frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\\ A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}\\ A< \frac{9}{10}< 1\Rightarrow A< 1\)
LC
2
21 tháng 4 2019
A=\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+.........+\(\frac{1}{100^2}\)
A=\(\frac{1}{3^2}\)<\(\frac{1}{2.3}\)
\(\frac{1}{4^2}\)<\(\frac{1}{3.4}\)
\(\Rightarrow\)\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{100^2}\)<\(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
=>\(\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}\)< \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{99}-\frac{1}{100}\)
=> \(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+.....+\(\frac{1}{100^2}\)< \(\frac{1}{2}-\frac{1}{100}\)
=>A< \(\frac{1}{2}\)
21 tháng 4 2019
Ta có: \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)
Ta thấy: \(\frac{1}{3^2}< \frac{1}{2\cdot3};\frac{1}{4^2}< \frac{1}{3\cdot4};\frac{1}{5^2}< \frac{1}{4\cdot5}...\frac{1}{100^2}< \frac{1}{99\cdot100}\)
\(\Rightarrow A< \frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{100}\Rightarrow A< \frac{1}{2}\left(ĐPCM\right)\)
IA
5 tháng 4 2018
Ta có 1/22<1/1.2
1/32<1/2.3
1/42<1/3.4
................
1/8²<1/7.8
=>B<1/1.2+1/2.3+1/3.4+...+1/7.8
=>B<1-1/2+1/2-1/3+1/3-1/4+...+1/7-1/8
=>B<1-1/8
Vậy B < 1
DT
0
DT
2
2 tháng 5 2018
b=1/22+1/32+1/42+...+1/82<1/1.2+1/2.3+1/3.4+......+1/7.8
b=1-1/2+1/2-1/3+1/3-1/4+....+1/7-1/8
b=1-1/8
b=7/8
<=>b<1
k cho mink nha
18 tháng 3 2024
b=1/22+1/32+1/42+...+1/82<1/1.2+1/2.3+1/3.4+......+1/7.8
b=1-1/2+1/2-1/3+1/3-1/4+....+1/7-1/8
b=1-1/8
b=7/8
<=>b<1
owo
VV
1
19 tháng 10 2023
`#3107.101107`
\(A = 2 + 2^2 + 2^3 + ... + 2^{2020} + 2^{2021} + 2^{2022}\)
\(= (2 + 2^2) + (2^3 + 2^4) + ... + (2^{2021} + 2^{2022})\)
\(=2(1+2) + 2^3(1 + 2) + ... + 2^{2021}(1 + 2)\)
\(=(1 + 2)(2 + 2^3 + ... + 2^{2021})\)
\(= 3(2 + 2^3 + ... + 2^{2021})\)
Vì \(3(2 + 2^3 + ... + 2^{2021})\) \(\vdots\) \(3\)
`\Rightarrow A \vdots 3`
Vậy, `A \vdots 3.`
Ta có A = \(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{2021}\)
= \(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2021}}\)
=> 2A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}\)
=> 2A - A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2021}}\right)\)
=> A = \(\frac{1}{2}-\frac{1}{2^{2021}}< \frac{1}{2}\left(\text{ĐPCM}\right)\)