So sánh:
A = 1+2+22+23+...+22022 và B = 5. 22021
nhanh hộ mình với
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Ta có \(A=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\)
\(2A=1+\dfrac{2}{2}+\dfrac{3}{2^2}+...+\dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\)
\(2A-A=\left(1+\dfrac{2}{2}+\dfrac{3}{2^2}+...+\dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\right)-\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\right)\)\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\) - \(\dfrac{2023}{2^{2023}}\)
Đặt B = \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\)
2B = \(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\)
2B - B = \(\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\right)\)B = 2 - \(\dfrac{1}{2^{2022}}\)
Suy ra A = 2 - \(\dfrac{1}{2^{2022}}\) - \(\dfrac{2023}{2^{2023}}\) < 2
Vậy A < 2
\(A=\dfrac{1}{2}+\dfrac{2}{2^{2}}+\dfrac{3}{2^{3}}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\)
\(2A=1+\dfrac22+\dfrac3{2^2}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\\2A-A=\left(1+\dfrac22+\dfrac3{2^2}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\right)-\left(\dfrac12+\dfrac2{2^2}+\dfrac3{2^3}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\right)\\A=1+\dfrac12+\dfrac1{2^3}\ +\,.\!.\!.+\ \dfrac1{2^{2021}}+\dfrac1{2^{2022}}-\dfrac{2023}{2^{2023}}\\2\left(A+\dfrac{2023}{2^{2023}}\right)=2+1+\dfrac12+\dfrac1{2^2}\ +\,.\!.\!.+\ \dfrac1{2^{2020}}+\dfrac1{2^{2021}}\\A+\dfrac{2023}{2^{2023}}=2-\dfrac1{2^{2022}}\\A=2-\dfrac1{2^{2022}}+\dfrac{2023}{2^{2023}}<2\)
A=1/2+1/22+1/23+...+1/22020+1/22021 > B=1/3+1/4+1/5+13/60
a) A = 2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰²²
2A = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰²³
A = 2A - A
= (2 + 2² + 2³ + 2⁴ + ... + 2²⁰²³) - (2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰²²)
= 2²⁰²³ - 2⁰
= 2²⁰²³ - 1
Vậy A = B
b) A = 2021 . 2023
= (2022 - 1).(2022 + 1)
= 2022.(2022 + 1) - 2022 - 1
= 2022² + 2022 - 2022 - 1
= 2022² - 1 < 2022²
Vậy A < B
a) \(A=2+2^2+2^3+...+2^{2022}\)
\(2A=2.\left(2+2^2+2^3+...+2^{2022}\right)\)
\(2.A=2^2+2^3+2^4+...+2^{2023}\)
\(2A-A=\left(2^2+2^3+2^4+...+2^{2023}\right)-\left(2+2^2+2^3+...+2^{2022}\right)\)
\(A=2^{2023}-2\)
b) A + 2 = 2x
Hay \(\left(2^{2023}-2\right)+2=2^x\)
\(2^{2023}-2+2=2^x\)
\(2^{2023}=2^x\)
\(\Rightarrow x=2023\)
a, A = 21 + 22 + 23 + ...+ 22022
2A = 22 + 23 +...+ 22022 + 22023
2A - A = 22023 - 21
A = 22023 - 2
b, A + 2 = 2\(^x\) ⇒ 22023 - 2 + 2 = 2\(x\)
22023 = 2\(^x\)
2023 = \(x\)
a) Ta có A = 21 + 22 + 23 + ... + 22022
2A = 22 + 23 + 24 + ... + 22023
2A - A = ( 22 + 23 + 24 + ... + 22023 ) - ( 21 + 22 + 23 + ... + 22022 )
A = 22023 - 2
Lại có B = 5 + 52 + 53 + ... + 52022
5B = 52 + 53 + 54 + ... + 52023
5B - B = ( 52 + 53 + 54 + ... + 52023 ) - ( 5 + 52 + 53 + ... + 52022 )
4B = 52023 - 5
B = \(\dfrac{5^{2023}-5}{4}\)
b) Ta có : A + 2 = 2x
⇒ 22023 - 2 + 2 = 2x
⇒ 22023 = 2x
Vậy x = 2023
Lại có : 4B + 5 = 5x
⇒ 4 . \(\dfrac{5^{2023}-5}{4}\) + 5 = 5x
⇒ 52023 - 5 + 5 = 5x
⇒ 52023 = 5x
Vậy x = 2023
\(A=2+2^2+2^3+...+2^{2021}\)
=>\(2A=2^2+2^3+2^4+...+2^{2022}\)
=>\(2A-A=2^2+2^3+...+2^{2021}+2^{2022}-2-2^2-2^3-...-2^{2021}\)
=>\(A=2^{2022}-2\)
=>A<B
Lời giải:
$C=1-2+2^2-2^3+2^4-....+2^{2022}$
$2C=2-2^2+2^3-2^4+2^5-...+2^{2023}$
$\Rightarrow C+2C=(1-2+2^2-2^3+2^4-....+2^{2022})+(2-2^2+2^3-2^4+2^5-...+2^{2023})$
$\Rightarrow 3C=2^{2023}-1$
$\Rightarrow C=\frac{2^{2023}-1}{3}$
A = 101 - 99 + 97 - 95 + 93 -91 + ... + 5-3 + 1
A=( 101 - 99 ) + ( 97 - 95 ) +(93 - 91 ) + ... + (5 + 3 ) + 1
A = (2 + 2 + 2 + .. + 2 )+ 1
Xét dãy số: 101; 97; 93;...;5
Số số hạng của dãy số trên là
[ ( 101 - 3 ) : 2 + 1 ] : 2 = 25
tổng của dãy số A là
2x 25 + 1 = 51
Đáp số 51
A=887 .884 B=886.885
A= 884 . 886 + 884 B = 886 . 884 +886
Vì 884 < 886
⇒A < B
\(A=1+2+2^2+...+2^{2022}\)
\(\Rightarrow2A=2+2^2+...+2^{2023}\)
\(\Rightarrow2A-A=2^{2023}-1\)
\(\Rightarrow A=2^{2023}-1\)
\(\Rightarrow A< 2^{2023}=2^2.2^{2021}=4.2^{2021}< 5^{2021}\)
\(\Rightarrow A< B\)