(1/2^2-1)(1/3^2-1)(1/4^2-1)...(1/2023^2-1)
1 + 1/2 * (1 + 2) + 1/3 * (1 + 2 + 3) + 1/4 * (1 + 2 + 3 + 4) +...+ 1 2023 (1+2+...+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
tham khảo thôi nha
1 + 1/2 * (1 + 2) + 1/3 * (1 + 2 + 3) + 1/4 * (1 + 2 + 3 + 4) +...+ 1 2023 (1+2+...+2023)
2023-1/2*(1+2)-1/3*(1+2+3)-1/4*(1+2+3+4)-...-1/2022*(1+2+3+4+...+2022)
Tính: 1/(1+2+3) + 1/(1+2+3+4) + 1/(1+2+3+4+5) + ... + 1/(1+2+3+...+2023)
Lời giải:
Gọi tổng trên là $A$
$A=\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+....+\frac{1}{\frac{2023.2024}{2}}$
$=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2023.2024}$
$=2(\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{2024-2023}{2023.2024})$
$=2(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{2023}-\frac{1}{2024})$
$=2(\frac{1}{3}-\frac{1}{2024})=\frac{2021}{3036}$
1. a,23/27-(11/17-4/27) +(28/17) b,2/3 .7/9 +2/3 .2/9-2/9 c,7/3:5/11-1/3.11/5 d,(1+1/2) .(1+1/3) .(1+1/4).......(1+1/2023) e, (1-1/2).(1-1/3).....(1-1/2023) f,13/17.5/11-7/13.2/5+5/11.4/17-2/5.6/13 g, 1/2+1/3.1/4-1/5:1/6 2,so sánh n+2/n+3 và n+1/n+2 ( n là số tự nhiên)
1:
a: =23/27-11/17+4/27+28/17
=23/27+4/27+28/17-11/17
=1+1=2
b: \(=\dfrac{2}{3}\cdot\left(\dfrac{7}{9}+\dfrac{2}{9}\right)-\dfrac{2}{9}\)
=2/3-2/9
=6/9-2/9
=4/9
c: \(=\dfrac{11}{5}\cdot\dfrac{7}{3}-\dfrac{1}{3}\cdot\dfrac{11}{5}\)
=11/5(7/3-1/3)
=11/5*2
=22/5
d: \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2024}{2023}=\dfrac{2024}{2}=1012\)
e: \(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2022}{2023}=\dfrac{1}{2023}\)
c=(1/2^(2)-1)*(1/3^(2)-1)*(1/4^(2)-1)...(1/2023^(2)-1)
Lời giải:
\(C=(\frac{1}{2^2}-1)(\frac{1}{3^2}-1)(\frac{1}{4^2}-1)....(\frac{1}{2023^2}-1)\)
\(=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.\frac{1-4^2}{4^2}....\frac{1-2023^2}{2023^2}\)
\(=\frac{(2^2-1)(3^2-1)(4^2-1)....(2023^2-1)}{2^2.3^2.4^2....2023^2}\)
\(=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)....(2023-1)(2023+1)}{2^2.3^2.4^2....2023^2}\)
\(=\frac{1.3.2.4.3.5.....2022.2024}{(2.3.4...2023)(2.3.4...2023)}\)
\(=\frac{(1.2.3...2022)(3.4.5....2024)}{(2.3...2023)(2.3.4...2023)}\)
\(=\frac{1}{2023}.\frac{2024}{2}=\frac{1012}{2023}\)
thu gọn a=1/2-1/2^2+1/2^3-1/2^4+...+1/2^2023-1/2^2024
\(A=\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3-\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{2023}-\left(\dfrac{1}{2}\right)^{2024}\)
\(A=\dfrac{2}{2^2}-\dfrac{1}{2^2}+\dfrac{2}{2^4}-\dfrac{1}{2^4}+...+\dfrac{2}{2^{2024}}-\dfrac{1}{2^{2024}}\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+...+\dfrac{1}{2^{2024}}\)
\(A=\dfrac{2^{2022}}{2^{2024}}+\dfrac{2^{2020}}{2^{2024}}+\dfrac{2^{2018}}{2^{2024}}+...+\dfrac{1}{2^{2024}}\)
\(2^2A=\dfrac{2^{2024}}{2^{2024}}+\dfrac{2^{2022}}{2^{2024}}+\dfrac{2^{2020}}{2^{2024}}+...+\dfrac{2^2}{2^{2024}}\)
\(\Rightarrow4A-A=3A=1-\dfrac{2}{2^{2024}}-\dfrac{1}{2^{2024}}\)
\(3A=1-\left(\dfrac{2}{2^{2024}}+\dfrac{1}{2^{2024}}\right)\)
\(3A=1-\dfrac{3}{2^{2024}}\)
\(A=\dfrac{1-\dfrac{3}{2^{2024}}}{3}\)
\(A=\dfrac{3\left(\dfrac{1}{3}-\dfrac{1}{2^{2024}}\right)}{3}\)
\(A=\dfrac{1}{3}-\dfrac{1}{2^{2024}}\)
giúp mk vs các bn. chiều nay mk phải nộp r
CMR1×2-1/2!+2×3-1/2!+3×4-1/4!+...+2023×2024/2024!<2
TH1
42:x=6
x= 42 :6
X= 7
TH 2
36:x = 6
X = 36: 6
X= 6
B=1/2-1/2^2+1/2^3-1/2^4+..... -1/2^2022+1/2^2023
Thu gọn B
\(B=\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3-\left(\dfrac{1}{2}\right)^4+...-\dfrac{1}{2022}+\dfrac{1}{2023}\\ \Rightarrow B=\dfrac{2}{2^2}-\dfrac{1}{2^2}+\dfrac{2}{2^4}-\dfrac{1}{2^4}+...+\dfrac{2}{2^{2024}}-\dfrac{1}{2^{2024}}\)
\(\Rightarrow B=\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+...+\dfrac{1}{2^{2024}}\)
\(\Rightarrow B=\dfrac{2^{2022}}{2^{2024}}+\dfrac{2^{2020}}{2^{2024}}+...+\dfrac{1}{2^{2024}}\\ \Rightarrow2^2B=\dfrac{2^{2024}}{2^{2024}}+\dfrac{2^{2022}}{2^{2024}}+...+\dfrac{2^2}{2^{2024}}\)
\(\Rightarrow4B-B=\dfrac{2}{2^{2024}}-\dfrac{1}{2^{2024}}\\ \Rightarrow3B=1-\left(\dfrac{2}{2^{2024}}+\dfrac{1}{2^{2024}}\right)\)
\(\Rightarrow3B=1-\dfrac{3}{2^{2024}}\\ \Rightarrow B=\dfrac{1-\dfrac{3}{2^{2024}}}{3}\)
\(\Rightarrow B=\dfrac{3\left(\dfrac{1}{3}-\dfrac{1}{2^{2024}}\right)}{3}\\ B=\dfrac{1}{3}-\dfrac{1}{2^{2024}}\)