tính c= 1+(-3)+(-3)^2+(-3)^3+...+(-3)^2022
QV
Những câu hỏi liên quan
Tính nhanh C=1/1*2*3+1/2*3*4+1/3*4*5+........+1/2021*2022*2023
Ta có: C = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/2021.2022.2023
=> C = 1/2. (3-1/1.2.3 + 4-2/2.3.4 + 5-3/3.4.5 + ... + 2023-2021/2021.2022.2023
=> C = 1/2. (1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 + ... + 1/2021.2022 - 1/2022.2023)
=> C = 1/2. (1/1.2 - 1/2022.2023)
- Phần còn lại bạn tự tính chứ số to quá
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c/m M=1/3+2/3^2+3/3^3+....+2022/3^2022 không là số nguyên
Tính giá trị biểu thức sau:
(1^1+2^2+3^3+......+2022^2022)(8^2-576:3^2)
giúp mik với ạ mình đang cần gấp.
\(\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(8^2-576:3^2\right)\)
\(=\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(64-576:3^2\right)\)
\(=\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(64-64\right)\)
\(=\left(1^1+2^2+3^3+4^4+2022^{2022}\right).0\)
\(=0\)
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Ta có :
82 - 576 : 32
= 64 - 576 : 9
= 64 - 64
= 0
(11 + 22 + 33 + 44 +...+ 20222022) . 0
= 0
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Kết quả của phép nhân (x + 2022)(x - 1)là : A.x^2+ 2022x-1 B.x^2+2021x - 2022- C.x^2023x - 2022 D.x^2 - 2021x + 2022 Biểu thức thích hợp là là (a + b) (A^2- AB + B^2) =..... A.A^3 + B^3 B.( A + B)^3 C. A^3 - B^3 D.(A-B)^3
\(1,\left(x+2022\right)\left(x-1\right)=x^2+2021x-2022\left(B\right)\\ 2,\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3\left(A\right)\)
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Tính
a. 3 + 3^3 + 3^5 + 3^7 +......+ 3^51 + 3^53
b. 1 + 4^2 + 4^4 +......+ 4^2020 + 4^2022
c. Bik : 1^3 + 2^3 + 3^3 +...+ 9^3 = 2025
Hãy tính E = 2^3 + 4^3 + 6^3 +...+ 18^3
so sánh b=1/2022+2/2021+3/2020+...+2021/2+2022/1 VÀ c=1/2+1/3+1/4+...+1/2022+1/2023
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
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1) Tính tổng C left(1-dfrac{1}{2}right).left(1-dfrac{1}{3}right).left(1-dfrac{1}{4}right).....left(1-dfrac{1}{2022}right)
2) Cho tổng A dfrac{1}{3} - dfrac{2}{3^2} + dfrac{3}{3^3} - dfrac{4}{3^4} +...+ dfrac{99}{3^{99}} - dfrac{100}{3^{100}}. Chứng tỏ rằng A dfrac{3}{16}
Đọc tiếp
1) Tính tổng C = \(\left(1-\dfrac{1}{2}\right)\).\(\left(1-\dfrac{1}{3}\right)\).\(\left(1-\dfrac{1}{4}\right)\).....\(\left(1-\dfrac{1}{2022}\right)\)
2) Cho tổng A = \(\dfrac{1}{3}\) - \(\dfrac{2}{3^2}\) + \(\dfrac{3}{3^3}\) - \(\dfrac{4}{3^4}\) +...+ \(\dfrac{99}{3^{99}}\) - \(\dfrac{100}{3^{100}}\). Chứng tỏ rằng A < \(\dfrac{3}{16}\)
1) Ta có
\(C=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2022}\right)\)
\(C=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2021}{2022}\)
\(C=\dfrac{1}{2022}\)
2) \(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)
\(\Rightarrow3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)
\(\Rightarrow4A=A+3A\) \(=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
\(\Rightarrow12A=3.4A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)
\(\Rightarrow16A=12A+4A=\left(3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\right)+\left(1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\right)\)
\(=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}\) \(< 3\). Từ đó suy ra \(A< \dfrac{3}{16}\)
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1*2022+2*2021+3*2020+.........................2022*1 tính
Tính các tổng sau B = 1.3 + 2+3 mũ2 + 3 . 3 mũ 2 + ... + 2022 . 3 mũ 2022 + 2023 . 3 mũ 2023