ND
1
K
Khách
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1 tháng 11 2023
a:
Số số hạng trong dãy M là:
(1002-12):10+1=100(số)
=>Sẽ có 50 cặp (1002;992); (982;972);....;(22;12) có hiệu bằng 10
\(M=1002-992+982-972+...+22-12\)
\(=\left(1002-992\right)+\left(982-972\right)+...+\left(22-12\right)\)
\(=10+10+...+10\)
=10*50=500
b: \(N=\left(202+182+...+42+22\right)-\left(192+172+...+32+12\right)\)
\(=\left(202-192\right)+\left(182-172\right)+...+\left(22-12\right)\)
=10+10+...+10
=10*10=100
NT
1
DS
27 tháng 6 2016
S=1/30+1/31+1/32+1/33+...+1/59+1/60
S có 31 phân số,ta thấy:
1/30>1/62 1/31>1/62 1/32>1/62 ............ 1/60>1/62
Vậy:
S>31.1/62
S>31/62
S>1/2
Vậy S>1/2
Chúc em học tốt^^
H9
HT.Phong (9A5)
CTVHS
22 tháng 6 2023
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
LV
0
JB
1
24 tháng 2 2017
ta có \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right).......\left(\frac{1}{10}-1\right)\)
\(A=-\left(\frac{1}{2}.\frac{2}{3}.....\frac{9}{10}\right)\)
\(A=-\frac{1}{10}\)
vi\(-\frac{1}{10}>-\frac{1}{9}\)
do đó A>\(\frac{-1}{9}\)
T
2
A
24 tháng 2 2019
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(A=\left(\frac{2}{2}-\frac{1}{2}\right)\left(\frac{3}{3}-\frac{1}{3}\right)...\left(\frac{19}{19}-\frac{1}{19}\right)\left(\frac{20}{20}-\frac{1}{20}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{18}{19}.\frac{19}{20}\)
\(A=\frac{1.2.3...18.19}{2.3.4...19.20}\)
\(A=\frac{1}{20}\Leftrightarrow A>\frac{1}{21}\)
24 tháng 2 2019
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{20}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}......\frac{19}{20}=\frac{1}{20}>\frac{1}{21}\)
\(\text{Vậy: A lớn hơn 1/21}\)
LP
7
26 tháng 8 2016
A = 1/1×2 + 1/2×3 + 1/3×4 + .. + 1/99×100
A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100
A = 1 - 1/100 < 1
PS
26 tháng 8 2016
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=1-\frac{1}{100}< 1\)
=> ĐPCM
19 tháng 8 2023
Each term of S is n!(n2 + n + 1) = n![n(n + 1) + 1] = n(n + 1)n! + n!
By definition, n(n + 1)n! + n! = n! + n(n + 1)!
Therefore, S can be simplified as
1! + 1.2! + 2! + 2.3! + ... + 100! + 100.101!
So \(\dfrac{S+1}{101!}=\dfrac{1+1!+1\cdot2!+2!+2\cdot3!+...+100!+100\cdot101!}{101!}\)
\(=\dfrac{2!+1\cdot2!+2!+2\cdot3!+3!+...+100!+100\cdot101!}{101!}\)
\(=\dfrac{3!+2\cdot3!+3!+...+100!+100\cdot101!}{101!}\)
\(=\dfrac{4!+3\cdot4!+4!+...+100!+100\cdot101!}{101!}\)
\(=...\)
\(=\dfrac{100!+99\cdot100!+100!+100\cdot101!}{101!}\)
\(=\dfrac{101!+100\cdot101!}{101!}\)
\(=1+100=101\)
Hence, \(\dfrac{S+1}{101!}=101\)
Ta có: \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(A=\frac{-3}{2^2}.\frac{-8}{3^2}.\frac{-15}{4^2}...\frac{-9900}{100^2}\)
\(A=\frac{\left(-1\right).3}{2^2}.\frac{\left(-2\right).4}{3^2}.\frac{\left(-3\right).5}{4^2}...\frac{\left(-99\right).101}{100^2}\)
\(A=\cdot\frac{\left(-1\right).\left(-2\right).\left(-3\right)...\left(-99\right)}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}\)
\(A=\left(-\frac{1}{100}\right).\frac{101}{2}\)
\(A=-\frac{101}{200}\)