Tính tổng \(S=2C^2_{2024}+3.2.9C^3_{2024}+4.3.9^2C^4_{2024}+...+2023.2024.9^{2022}.C^{2024}_{2024}\).
Tính: \(S=C^0_{2024}+\dfrac{1}{2}.C^2_{2024}+\dfrac{1}{3}.C^4_{2024}+\dfrac{1}{4}.C^6_{2024}+...+\dfrac{1}{1013}.C^{2024}_{2024}\)
Tính: \(S=C^0_{2024}+\dfrac{1}{2}.C^2_{2024}+\dfrac{1}{3}.C^4_{2024}+\dfrac{1}{4}.C^6_{2024}+...+\dfrac{1}{1013}.C^{2024}_{2024}\)
Tính: \(S=C^0_{2024}+\dfrac{1}{2}.C^2_{2024}+\dfrac{1}{3}.C^4_{2024}+\dfrac{1}{4}.C^6_{2024}+...+\dfrac{1}{1013}.C^{2024}_{2024}\)
\(S=C^0_{2024}+\dfrac{1}{2}C^2_{2024}+\dfrac{1}{3}C^4_{2024}+\dfrac{1}{4}C^6_{2024}+...+\dfrac{1}{1013}C^{2024}_{2024}\)
Ta có :
\(\dfrac{1}{k+1}C^{2k-1}_n=\dfrac{1}{k+1}.\dfrac{n!}{\left(2k-1\right)!\left(n-2k+1\right)!}\)
\(=\dfrac{1}{n+1}.\dfrac{\left(n+1\right)!}{2k!\left[\left(n+1\right)-2k\right]!}\)
\(=\dfrac{1}{n+1}C^{2k}_{n+1}\)
\(\Rightarrow S_n=\dfrac{1}{n+1}\Sigma^{2k}_{k=0}C^{2k}_{n+1}=\dfrac{1}{n+1}\left(\Sigma^{2k}_{k=0}C^{2k-1}_{n+1}-C^0_{n+1}\right)=\dfrac{2^{2n-1}-1}{n+1}\)
\(\Rightarrow S=\dfrac{2^{2025}-1}{1013}\)
Tính: \(S=C^0_{2024}+\dfrac{1}{2}.C^2_{2024}+\dfrac{1}{3}.C^4_{2024}+\dfrac{1}{4}.C^6_{2024}+...+\dfrac{1}{1013}.C^{2024}_{2024}\)
S = C₀₂₀₂₄ + 12.C₂₀₂₄ + 13.C₂₀₂₄ + 14.C₂₀₂₄ + ... + 11013.C₂₀₂₄
= (C₀₂₀₂₄ + C₂₀₂₄ + C₂₀₂₄ + C₂₀₂₄ + ... + C₂₀₂₄) + (C₂₀₂₄ + C₂₀₂₄ + C₂₀₂₄ + ... + C₂₀₂₄) + ... + (C₂₀₂₄)
= 11014.C₂₀₂₄
= 11014.
Tính: \(S=C^0_{2024}+\dfrac{1}{2}.C^2_{2024}+\dfrac{1}{3}.C^4_{2024}+\dfrac{1}{4}.C^6_{2024}+...+\dfrac{1}{1013}.C^{2024}_{2024}\)
Tính: \(S=C^0_{2024}+\dfrac{1}{2}.C^2_{2024}+\dfrac{1}{3}.C^4_{2024}+\dfrac{1}{4}.C^6_{2024}+...+\dfrac{1}{1013}.C^{2024}_{2024}\)
Tính: \(S=C^0_{2024}+\dfrac{1}{2}.C^2_{2024}+\dfrac{1}{3}.C^4_{2024}+\dfrac{1}{4}.C^6_{2024}+...+\dfrac{1}{1013}.C^{2024}_{2024}\)
Giúp mình với!!!
So sánh A=\(\dfrac{2024^{2023}+1}{2024^{2024}+1}\) và B=\(\dfrac{2024^{2022}+1}{2024^{2023}+1}\)
Cám ơn các bạn!
\(A=\dfrac{2024^{2023}+1}{2024^{2024}+1}\)
\(2024A=\dfrac{2024^{2024}+2024}{2024^{2024}+1}=\dfrac{\left(2024^{2024}+1\right)+2023}{2024^{2024}+1}=\dfrac{2024^{2024}+1}{2024^{2024}+1}+\dfrac{2023}{2024^{2024}+1}=1+\dfrac{2023}{2024^{2024}+1}\)
\(B=\dfrac{2024^{2022}+1}{2024^{2023}+1}\)
\(2024B=\dfrac{2024^{2023}+2024}{2024^{2023}+1}=\dfrac{\left(2024^{2023}+1\right)+2023}{2024^{2023}+1}=\dfrac{2024^{2023}+1}{2024^{2023}+1}+\dfrac{2023}{2024^{2023}+1}=1+\dfrac{2023}{2024^{2023}+1}\)
Vì \(2024>2023=>2024^{2024}>2024^{2023}\)
\(=>2024^{2024}+1>2024^{2023}+1\)
\(=>\dfrac{2023}{2024^{2023}+1}>\dfrac{2023}{2024^{2024}+1}\)
\(=>A< B\)
\(#PaooNqoccc\)
\(P\left(x\right)\)=\(x^{2023}-2024.x^{2022}+2024.x^{2021}-2024.x^{2020}+.....+2024.x-1\)
tính P ( 2023)
Giải nhanh giúp mik ạ !! đang cânf gấp O(∩_∩)O
Với x = 2023
<=> x + 1 = 2024
Khi đó P(2023) = x2023 - (x + 1).x2022 + ... + (x + 1).x - 1
= x2023 - x2023 - x2022 + .. + x2 + x - 1
= x - 1 = 2023 - 1 = 2022
tính nhanh
(2022 x 2023 + 2024 x 21 + 2002 ) :( 2024 x 2023 - 2022 x 2023 )