B=1/2-1/2^2+1/2^3-1/2^4+..... -1/2^2022+1/2^2023
Thu gọn B
Những câu hỏi liên quan
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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2023-1/2*(1+2)-1/3*(1+2+3)-1/4*(1+2+3+4)-...-1/2022*(1+2+3+4+...+2022)
B= 1/2 + 1/3 - 1/4 +...- 1/2022 + 1/2023 C= 1/1012 + 1/1013+...+ 1/2022 + 1/2023
Tính: B-C
mọi người sửa nhanh giúp mik vs ạ
2. Cho:
B= 1 - 1/2 + 1/3 - 1/4 +...+ 1/2021 - 1/2022 + 1/2023 C= 1/1012 + 1/1013 + 1/1014 +...+ 1/2021 + 1/2022 + 1/2023
Tính: B-C
1. Rút gọn các biểu thức sau:
A= 1 - 2 + 3 - 4 + 5 - 6 + ... + 2021 - 2022 + 2023
A=(-1)+(-1)+...+(-1)+2023
=2023-1011
=1012
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thu gọn a=1/2-1/2^2+1/2^3-1/2^4+...+1/2^2023-1/2^2024
\(A=\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3-\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{2023}-\left(\dfrac{1}{2}\right)^{2024}\)
\(A=\dfrac{2}{2^2}-\dfrac{1}{2^2}+\dfrac{2}{2^4}-\dfrac{1}{2^4}+...+\dfrac{2}{2^{2024}}-\dfrac{1}{2^{2024}}\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+...+\dfrac{1}{2^{2024}}\)
\(A=\dfrac{2^{2022}}{2^{2024}}+\dfrac{2^{2020}}{2^{2024}}+\dfrac{2^{2018}}{2^{2024}}+...+\dfrac{1}{2^{2024}}\)
\(2^2A=\dfrac{2^{2024}}{2^{2024}}+\dfrac{2^{2022}}{2^{2024}}+\dfrac{2^{2020}}{2^{2024}}+...+\dfrac{2^2}{2^{2024}}\)
\(\Rightarrow4A-A=3A=1-\dfrac{2}{2^{2024}}-\dfrac{1}{2^{2024}}\)
\(3A=1-\left(\dfrac{2}{2^{2024}}+\dfrac{1}{2^{2024}}\right)\)
\(3A=1-\dfrac{3}{2^{2024}}\)
\(A=\dfrac{1-\dfrac{3}{2^{2024}}}{3}\)
\(A=\dfrac{3\left(\dfrac{1}{3}-\dfrac{1}{2^{2024}}\right)}{3}\)
\(A=\dfrac{1}{3}-\dfrac{1}{2^{2024}}\)
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giúp mk vs các bn. chiều nay mk phải nộp r
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B = 1×3+2×3(mũ 2)+3×3(mũ 3)+...+2022×3(mũ 2022)+2023×3(mũ 2023)
\(3B=1.3^2+2.3^3+3.3^4+...+2022.3^{2023}+2023.3^{2024}\)
\(2B=3B-B=-3-3^2-3^3-...-3^{2023}+2023.3^{2024}\)
\(2B=2023.3^{2024}-\left(3+3^2+3^3+...+3^{2023}\right)\)
Đặt
\(C=3+3^2+3^3+...+3^{2023}\)
\(3C=3^2+3^3+3^4+...+3^{2024}\)
\(2C=3C-C=3^{2024}-3\Rightarrow C=\dfrac{3^{2024}-3}{2}\)
\(\Rightarrow2B=2023.3^{2024}-\dfrac{3^{2024}-3}{2}=\)
\(=\dfrac{2.2023.3^{2024}-3^{2024}+3}{2}=\dfrac{4045.3^{2024}+3}{2}\)
\(\Rightarrow B=\dfrac{4045.3^{2024}+3}{4}\)
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a)2022.2023-2022/2021.2022+2022
b)1999.2000-1/1998.1997+3997
c)(1-1/2).(1-1/3).(1-1/4).(1-1/5)...(1-1/2022).(1-1/2023)
help me
tui làm được câu c thui
c) (1-1/2).(1-1/3).(1-1/4).(1-1/5)...(1-1/2022).(1-1/2023)
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2^2+6^3+12^4+......................................+[(k+1)*k]^(k+1)+.......................+(2022*2023)^2023
@Bé Bin bạn có biết cách giải không zậy
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