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NS
2
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
AT
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 9 2023
A = \(\dfrac{1}{1+2+3}\)+\(\dfrac{1}{1+2+3+4}\)+...+ \(\dfrac{1}{1+2+...+2004}\)+ \(\dfrac{2}{2025}\)
A = \(\dfrac{1}{\left(1+3\right).3:2}\)+\(\dfrac{1}{\left(4+1\right).4:2}\)+...+ \(\dfrac{1}{\left(2024+1\right).2024:2}\)+\(\dfrac{2}{2025}\)
A = \(\dfrac{2}{3.4}\)+\(\dfrac{2}{4.5}\)+...+\(\dfrac{2}{2024.2025}\)+ \(\dfrac{2}{2025}\)
A = 2.(\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\)+...+ \(\dfrac{1}{2024.2025}\)) + \(\dfrac{2}{2025}\)
A = 2.(\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+...+ \(\dfrac{1}{2024}\) - \(\dfrac{1}{2025}\)) + \(\dfrac{2}{2025}\)
A = 2.(\(\dfrac{1}{3}\) - \(\dfrac{1}{2025}\)) + \(\dfrac{2}{2025}\)
A = \(\dfrac{2}{3}\) - \(\dfrac{2}{2025}\) + \(\dfrac{2}{2025}\)
A = \(\dfrac{2}{3}\)
MM
3
3 tháng 5 2023
\(A=\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3-\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{2023}-\left(\dfrac{1}{2}\right)^{2024}\)
\(A=\dfrac{2}{2^2}-\dfrac{1}{2^2}+\dfrac{2}{2^4}-\dfrac{1}{2^4}+...+\dfrac{2}{2^{2024}}-\dfrac{1}{2^{2024}}\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+...+\dfrac{1}{2^{2024}}\)
\(A=\dfrac{2^{2022}}{2^{2024}}+\dfrac{2^{2020}}{2^{2024}}+\dfrac{2^{2018}}{2^{2024}}+...+\dfrac{1}{2^{2024}}\)
\(2^2A=\dfrac{2^{2024}}{2^{2024}}+\dfrac{2^{2022}}{2^{2024}}+\dfrac{2^{2020}}{2^{2024}}+...+\dfrac{2^2}{2^{2024}}\)
\(\Rightarrow4A-A=3A=1-\dfrac{2}{2^{2024}}-\dfrac{1}{2^{2024}}\)
\(3A=1-\left(\dfrac{2}{2^{2024}}+\dfrac{1}{2^{2024}}\right)\)
\(3A=1-\dfrac{3}{2^{2024}}\)
\(A=\dfrac{1-\dfrac{3}{2^{2024}}}{3}\)
\(A=\dfrac{3\left(\dfrac{1}{3}-\dfrac{1}{2^{2024}}\right)}{3}\)
\(A=\dfrac{1}{3}-\dfrac{1}{2^{2024}}\)
PT
2
ND
Nguyễn Đức Trí
VIP
15 tháng 7 2023
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2023}\right)\left(1-\dfrac{1}{2024}\right)\)
=\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2022}{2023}.\dfrac{2023}{2024}=\dfrac{1}{2024}\)
H9
HT.Phong (9A5)
CTVHS
15 tháng 7 2023
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2024}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot...\cdot\dfrac{2023}{2024}\)
\(=\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot2023}{2\cdot3\cdot4\cdot5\cdot...\cdot2024}\)
\(=\dfrac{1}{2024}\)
ND
Nguyễn Đăng Nhân
CTVHS
30 tháng 9 2023
\(S=1+3^2+3^4+...+3^{2022}\)
\(3^2S=9S=3^2+3^4+3^6+...+3^{2024}\)
\(S=\dfrac{9S-S}{8}=\left(3^{2024}-1\right):8\)
d, không đáp án nào đúng
AH
Akai Haruma
Giáo viên
30 tháng 9 2023
Lời giải:
$S=1+3^2+3^4+....+3^{2022}$
$9S=3^2S=3^2+3^4+3^6+...+3^{2024}$
$\Rightarrow 9S-S=3^{2024}-1$
$\Rightarrow S=\frac{3^{2024}-1}{8}$
Đáp án D.
LC
26 tháng 2 2017
A=(1+1+....+1)-(1/2 +1/4 +...+1/1024)
A=10-(1/2+ 1/4+...+1/1024)
ta có B=1/2+1/4+..1/1024
2B=1+1/2+....+1/512
2B-B=(1+1/2+...+1/512)-(1/2+1/4+...+1/1024)
B=1-1/1024=1023/1024
A=10-1023/1024=9217/1024
AV
1
PV
1
23 tháng 3 2023
P=[(1-2)+(-3+4)+(5-6)+(-7+8)+...+(993-994)+(-995+996)]+997
P=[(-1)+1+(-1)+1+...+(-1)+1+(-1)+1]+997
P= 0 +0 +...+ 0 +997
P=997