Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có \(\frac{a}{a^2}=\frac{a^2}{a^3}=...=\frac{a^{2020}}{a^{2021}}=\frac{a+a^2+....+a^{2020}}{a^2+a^3+...+a^{2021}}\)
=> \(\frac{a}{a^2}=\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\)
=> \(\left(\frac{a}{a^2}\right)^{2020}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)
=> \(\frac{a}{a^2}.\frac{a}{a^2}...\frac{a}{a^2}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)(2020 thừa số \(\frac{a}{a^2}\))
=> \(\frac{a}{a^2}.\frac{a^2}{a^3}...\frac{a^{2020}}{a^{2021}}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)(Vì \(\frac{a}{a^2}=\frac{a^2}{a^3}=...=\frac{a^{2020}}{a^{2021}}\))
=> \(\frac{a}{a^{2021}}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)(đpcm)
a) Ta có:
2A=2.(12+122+123+...+122020+122021)2�=2.12+122+123+...+122 020+122 021
2A=1+12+122+123+...+122019+1220202�=1+12+122+123+...+122 019+122 020
Suy ra: 2A−A=(1+12+122+123+...+122019+122020)2�−�=1+12+122+123+...+122 019+122 020
−(12+122+123+...+122020+122021)−12+122+123+...+122 020+122 021
Do đó A=1−122021<1�=1−122021<1.
Lại có B=13+14+15+1360=20+15+12+1360=6060=1�=13+14+15+1360=20+15+12+1360=6060=1.
Vậy A < B.
S = \(\left(1+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2020}\right)\)
= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2020}\right)\)
= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1010}\right)\)
= \(\dfrac{1}{1011}+\dfrac{1}{1012}+...+\dfrac{1}{2021}\)
B = \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\) + ... + \(\dfrac{1}{3^{2020}}\) + \(\dfrac{1}{3^{2021}}\) < \(\dfrac{1}{2}\)
3.B = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\)+ ... + \(\dfrac{1}{3^{2019}}\) + \(\dfrac{1}{3^{2020}}\)
3B - B = 1+\(\dfrac{1}{3}\)+ \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{3^{2019}}\) + \(\dfrac{1}{3^{2020}}\) - (\(\dfrac{1}{3}\)+ \(\dfrac{1}{3^2}\)+ ... + \(\dfrac{1}{3^{2020}}\)+\(\dfrac{1}{3^{2021}}\))
2B = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\)+...+ \(\dfrac{1}{3^{2019}}\) + \(\dfrac{1}{3^{2020}}\) - \(\dfrac{1}{3}\) - \(\dfrac{1}{3^2}\)- ...- \(\dfrac{1}{3^{2020}}\)-\(\dfrac{1}{3^{2021}}\)
2B = (1 - \(\dfrac{1}{3^{2021}}\)) + (\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\)) + (\(\dfrac{1}{3^2}\) - \(\dfrac{1}{3^2}\)) +...+ (\(\dfrac{1}{3^{2020}}\) - \(\dfrac{1}{3^{2020}}\))
2B = 1 - \(\dfrac{1}{3^{2021}}\)
B = (1 - \(\dfrac{1}{3^{2021}}\)) : 2
B = \(\dfrac{1}{2}\) - \(\dfrac{1}{2.3^{2021}}\) < \(\dfrac{1}{2}\) (đpcm)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{x\cdot\left(x+1\right)}=\frac{2020}{2021}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2020}{2021}\)
\\(1-\frac{1}{x+1}=\frac{2020}{2021}\)
\(\frac{1}{x+1}=1-\frac{2020}{2021}\)
\(\frac{1}{x+1}=\frac{1}{2021}\)
\(\Rightarrow x+1=2021\)
\(x=2021-1\)
\(x=2020\)
đk: \(x\ne\left\{0;-1\right\}\)
Ta có: \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{2020}{2021}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2020}{2021}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{2020}{2021}\)
\(\Leftrightarrow\frac{x}{x+1}=\frac{2020}{2021}\)
\(\Leftrightarrow2021x=2020x+2020\)
\(\Rightarrow x=2020\)
\(B=\dfrac{\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}}{\dfrac{3}{2020}+\dfrac{3}{2021}-\dfrac{3}{2022}}-1=\dfrac{\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}}{3\left(\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}\right)}-1=\dfrac{1}{3}-1=-\dfrac{2}{3}\)
\(B=\dfrac{\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}}{\dfrac{3}{2020}+\dfrac{3}{2021}-\dfrac{3}{2022}}-1=\dfrac{\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}}{3\left(\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}\right)}-1=\dfrac{1}{3}-1=\dfrac{1}{3}-\dfrac{3}{3}=-\dfrac{2}{3}\)