(x+1/2)^2024=2-(1+1/2+1/2^2+1/2^3+...+ 1/2^2024)
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NT
1
28 tháng 9 2023
1+1/2.(1+2)+1/3.(1+2+3)+1/4.(1+2+3+4)+...+1/2023.(1+2+3+...+2023)
=1+1/2.(1+2).2/2+1/3.(1+3).3/2+1/4.(1+4).4/2+...+1/2023.(1+2+3+...+2023).2023/2
=2/2+3/2+4/2+...+2023/2
=2+3+4+...+2023/2
=2025.2022/2/2
=1023637,5
NS
2
NA
9 tháng 1 2024
a, 2\(^3\) . x + 2005\(^0\) . x = 994-15:3+1\(^{2025}\)
8 .x + 1 . x = 990
x . [ 8 +1 ] = 990
x . 9 = 990
x = 990 : 9
x = 110
NL
Nguyễn Lê Phước Thịnh
CTVHS
10 tháng 1 2024
a: \(\left(2^3\right)^{1^{2005}}\cdot x+2005^0\cdot x=9915:3+1^{2025}\)
=>\(8\cdot x+1\cdot x=3305+1\)
=>\(9x=3306\)
=>\(x=\dfrac{3306}{9}=\dfrac{1102}{3}\)
b: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
=>\(2^x+2^x\cdot2+2^x\cdot4+2^x\cdot8=480\)
=>\(2^x\left(1+2+4+8\right)=480\)
=>\(2^x\cdot15=480\)
=>\(2^x=32\)
=>\(2^x=2^5\)
=>x+5
10 tháng 9 2023
\(S=C^0_{2024}+\dfrac{1}{2}C^2_{2024}+\dfrac{1}{3}C^4_{2024}+\dfrac{1}{4}C^6_{2024}+...+\dfrac{1}{1013}C^{2024}_{2024}\)
Ta có :
\(\dfrac{1}{k+1}C^{2k-1}_n=\dfrac{1}{k+1}.\dfrac{n!}{\left(2k-1\right)!\left(n-2k+1\right)!}\)
\(=\dfrac{1}{n+1}.\dfrac{\left(n+1\right)!}{2k!\left[\left(n+1\right)-2k\right]!}\)
\(=\dfrac{1}{n+1}C^{2k}_{n+1}\)
\(\Rightarrow S_n=\dfrac{1}{n+1}\Sigma^{2k}_{k=0}C^{2k}_{n+1}=\dfrac{1}{n+1}\left(\Sigma^{2k}_{k=0}C^{2k-1}_{n+1}-C^0_{n+1}\right)=\dfrac{2^{2n-1}-1}{n+1}\)
\(\Rightarrow S=\dfrac{2^{2025}-1}{1013}\)
M
10 tháng 9 2023
S = C₀₂₀₂₄ + 12.C₂₀₂₄ + 13.C₂₀₂₄ + 14.C₂₀₂₄ + ... + 11013.C₂₀₂₄
= (C₀₂₀₂₄ + C₂₀₂₄ + C₂₀₂₄ + C₂₀₂₄ + ... + C₂₀₂₄) + (C₂₀₂₄ + C₂₀₂₄ + C₂₀₂₄ + ... + C₂₀₂₄) + ... + (C₂₀₂₄)
= 11014.C₂₀₂₄
= 11014.
Đặt \(A=1+\frac12+\frac{1}{2^2}+\cdots+\frac{1}{2^{2024}}\)
=>\(2A=2+1+\frac12+\cdots+\frac{1}{2^{2023}}\)
=>\(2A-A=2+1+\frac12+\cdots+\frac{1}{2^{2023}}-1-\frac12-\frac{1}{2^2}-\cdots-\frac{1}{2^{2024}}\)
=>\(A=2-\frac{1}{2^{2024}}\)
\(\left(x+\frac12\right)^{2024}=2-\left(1+\frac12+\frac{1}{2^2}+\cdots+\frac{1}{2^{2024}}\right)\)
=>\(\left(x+\frac12\right)^{2024}=2-\left(2-\frac{1}{2^{2024}}\right)=\frac{1}{2^{2024}}=\left(\frac12\right)^{2024}\)
=>\(\left[\begin{array}{l}x+\frac12=\frac12\\ x+\frac12=-\frac12\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac12-\frac12=0\\ x=-\frac12-\frac12=-\frac22=-1\end{array}\right.\)