A=1. 2+ 2.3+3. 4+......+48 .49
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25 tháng 8 2023
a: \(A=\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)+1\)
\(=\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}+\dfrac{100}{100}\)
\(=100\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)\)=100B
=>B/A=1/100
b: \(A=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\left(1\right)\)
\(=\dfrac{50}{49}+\dfrac{50}{48}+....+\dfrac{50}{2}+\dfrac{50}{50}\)
\(=50\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)
\(B=\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{49}+\dfrac{2}{50}\)
\(=2\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)\)
=>A/B=25
MV
23 tháng 5 2017
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\\ 2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\\ 2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\\ A=1-\dfrac{1}{2^{100}}\)
MV
23 tháng 5 2017
\(E=\dfrac{3^2}{2\cdot4}+\dfrac{3^2}{4\cdot6}+...+\dfrac{3^2}{198\cdot200}\\ =3^2\cdot\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{198\cdot200}\right)\\ =9\cdot\dfrac{1}{2}\cdot\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{198\cdot200}\right)\\ =\dfrac{9}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{198}-\dfrac{1}{200}\right)\\ =\dfrac{9}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{200}\right)\\ =\dfrac{9}{2}\cdot\dfrac{99}{200}\\ =\dfrac{891}{400}\)
18 tháng 5 2022
\(B=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(B=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\dfrac{49}{1}\)
\(B=\left(\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\right)+1\)
\(B=\dfrac{50}{50}+\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\)
\(B=50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}}{50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)}=\dfrac{1}{50}\)
LL
17 tháng 5 2017
Xin lỗi! Cho mình hỏi cái:
Đề bạn viết có đúng hay không, nếu đúng thì mình ko giải được, còn nếu đề là thế này thì mình giải được:
Cho P = 1/2 + 1/3 + ... + 1/48 + 1/49 + 1/50 và Q = 1/49 + 2/48 + ... + 48/2 + 49/2
Tính P/Q
Cách làm như sau:
Ta có:
Q = 1/49 + 2/48 + ... + 48/2 + 49/1
= 50 - 49/49 + 50 - 48/48 + ... + 50 - 2/2 + 50 - 1/1
= 50/49 - 1 + 50/48 - 1 + ... + 50/2 - 1 + 50/1 - 1
= 50/49 + 50/48 + ... + 50/2 + 50/1 - (1 + 1 + ... + 1 + 1) (49 số hạng 1)
= 50/49 + 50/48 + ... + 50/2 + 50 - 49
= 50/2 + ... + 50/48 + 50/49 + 1
= 50/2 + ... + 50/48 + 50/49 + 50/50
= 50.(1/2 + 1/3 + ... + 1/48 + 1/49 + 1/50)
=> P/Q = (1/2 + 1/3 + ... + 1/48 + 1/49 + 1/50) / 50.(1/2 + 1/3 + ... + 1/48 + 1/49 + 1/50) = 1/50
Vậy P/Q = 1/50
26 tháng 1 2023
So sánh tổng : S = 1/5 + 1/9 + 1/10 + 1/41 + 1/42 với 1/2
NC
26 tháng 1 2023
S=
=50/50+50/49+50/48+...+50/2
=50.(1/50+1/49+1/48+...+1/4+1/3+1/2)
=50
P=
P=(1/49+1)+(2/48+1)+...+(48/2+1)+1
P= 50/49+50/48+....+50/2+50/50=1
vậy s/p = 1/50
\(A=1.2+2.3+...+48.49\)
\(\Rightarrow3A=1.2.3+2.3.3+...+48.49.3\)
\(=1.2.3+2.3.\left(4-1\right)+...+48.49.\left(50-47\right)\)
\(=1.2.3+2.3.4-1.2.3+....+48.49.50-47.48.49\)
\(=48.49.50\)
\(\Rightarrow A=\frac{48.49.50}{3}=39200\)
A = 1.2 + 2.3 + 3.4 + ... + 48.49
=> 3.A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 48.49.3
=> 3.A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 48.49.(50 - 47)
=> 3.A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 48.49.50 - 47.48.49
=> 3.A = 48.49.50
=> A = 16.49.50
=> A = 39200