so sánh A=3+3^2+3^3....+3^2021 và B=5.3^2022 - 15
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2021/2022=(2022-1)/2022=1-1/2022
2022/2023=(2023-1)/2023=1-1/2023
Do: 2022<2023
=>1/2022>1/2023
=>1-1/2022<1-1/2023
Vậy 2021/2022 < 2022/2023
A=1+3+32+33+.....+32021
-->3A=3(1+3+32+33+.....+32021)
-->3A=3+32+33+...+32022
-->3A-A=(3+32+33+....32022)-(1+3+32+33+.....+32021)
-->2A=32022-1
-->A=(32022-1):2
Vì (32022-1):2>(32022-1):2
-->A=B
B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + \(\dfrac{2022}{1}\)
B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + 2022
B = 1 + ( 1 + \(\dfrac{1}{2022}\)) + ( 1 + \(\dfrac{2}{2021}\)) + \(\left(1+\dfrac{3}{2020}\right)\)+ ... + \(\left(1+\dfrac{2021}{2}\right)\)
B = \(\dfrac{2023}{2023}\) + \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2021}\) + \(\dfrac{2023}{2020}\) + ...+ \(\dfrac{2023}{2}\)
B = 2023 \(\times\) ( \(\dfrac{1}{2023}\) + \(\dfrac{1}{2022}\) + \(\dfrac{1}{2021}\) + \(\dfrac{1}{2020}\)+ ... + \(\dfrac{1}{2}\))
Vậy B > C
a) \(A=1+2+2^2+2^3+...+2^{100}\) \(B=2^{201}\)
\(2A=2\left(1+2+2^2+2^3+...+2^{100}\right)\)
\(2A=2+2^2+2^3+2^4+...+2^{201}\)
\(2A-A=\left(2+2^2+2^3+2^4+...+2^{201}\right)-\left(1+2+2^2+2^3+...+2^{100}\right)\)
\(2A-A=2^{101}-1\)
\(A=2^{201}-1\)
Ta có 2201 > 2201 - 1 => B > A => 2201 > 1 + 2 + 22 + 23 +...+ 1100
Lời giải:
$A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2022}}$
$3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2021}}$
$\Rightarrow 3A-A=1-\frac{1}{3^{2022}}$
$\Rightarrow A=\frac{1}{2}-\frac{1}{2.3^{2022}}$
Xét hiệu:
$A-B=\frac{1}{2}-\frac{1}{2.3^{2022}}-(1-\frac{1}{3^{2021}})$
$=\frac{1}{3^{2021}}-\frac{1}{2.3^{2022}}-\frac{1}{2}$
$=\frac{5}{2.3^{2022}}-\frac{1}{2}$
$< \frac{1}{2}-\frac{1}{2}=0$
$\Rightarrow A< B$
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.
A=2+22+23+...+22021
2A=22+23+24+...+22022
2A-A=(22+23+24+...+22022)-(2+22+23+...+22021)
A=22022-2 mà B= 22022 nên A<B.