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NN
0
6 tháng 10 2018
\(\frac{49}{1}+\frac{48}{2}+\frac{47}{3}+...+\frac{2}{48}+\frac{1}{49}\)
\(=1+1+...+1+\frac{48}{2}+\frac{47}{3}+...+\frac{2}{48}+\frac{1}{49}\)(có 49 số 1)
\(=\left(1+\frac{48}{2}\right)+\left(1+\frac{47}{3}\right)+...+\left(1+\frac{2}{48}\right)+\left(1+\frac{1}{49}\right)+1\)
\(=\frac{50}{2}+\frac{50}{3}+...+\frac{50}{48}+\frac{50}{49}+\frac{50}{50}\)
\(=50\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\right)\)
Chúc bạn học tốt.
XN
1
5 tháng 11 2017
S4 = 12 + 22 + 32 + ... + 492 + 502
S4 = 1 + 2 ( 1 + 1 ) + 3 ( 2 + 1 ) + ... + 49 ( 48 + 1 ) + 50 ( 49 + 1 )
S4 = 1 + 1.2 + 2 + 2.3 + 3 + ... + 48 . 49 + 49 + 49 . 50 + 50
S4 = ( 1 + 2 + 3 + ... 49 + 50 ) + ( 1.2 + 2.3 + ... + 48 . 49 + 49 . 50 )
đặt A = 1 + 2 + 3 + ... 49 + 50
Ta tính được : A = 1275
đặt B = 1.2 + 2.3 + ... + 48 . 49 + 49 . 50
3B = 1.2.3 + 2.3.3 + ... + 48.49.3 + 49.50.3
3B = 1.2.3 + 2.3.(4-1) + ... + 48.49.(50-47) + 49.50.(51-48)
3B = 1.2.3 + 2.3.4 - 1.2.3 + ... + 48.49.50 - 47.48.49 + 49.50.51-48.49.50
3B = 49.50.51
B = 49.50.51 : 3 = 41650
=> S4 = 41650 + 1275 = 42925
S5 = 13 + 23 + 33 + ... 493 + 503
S5 = 1 + 22 ( 1 + 1 ) + 32 ( 2 + 1 ) + ... 492 ( 48 + 1 ) + 502 ( 49 + 1 )
S5 = 12 + 1.22 + 22 + 2.32 + 32 + ... + 48.492 + 492 + 49.502 + 502
S5 = ( 12 + 22 + 32 + ... + 492 + 502 ) + ( 1.22 + 2.32 + ... + 48.492 + 49.502 )
đặt Y = 12 + 22 + 32 + ... + 492 + 502
Y = 42925
đặt M = 1.22 + 2.32 + ... + 48.492 + 49.502
M = 1.2.(3-1) + 2.3.(4-1) + ... + 48.49.(50-1) + 49.50.(51-48)
M = (1.2.3+2.3.4+...+48.49.50+49.50.51)-(1.2+2.3+...+48.49+49.50)
đến đây đơn giản rồi
9 tháng 4 2017
Bài 1:
Ta có:
\(S=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\)
\(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{49}{1}\)
\(\Rightarrow\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{49}{1}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{\left(1+\dfrac{1}{49}\right)+\left(1+\dfrac{2}{48}\right)+...+\left(1+\dfrac{48}{2}\right)+1}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)
Vậy \(\dfrac{S}{P}=\dfrac{1}{50}\)
Bài 2:
Ta có:
\(S=\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}\)
\(=\dfrac{1}{5}+\left(\dfrac{1}{9}+\dfrac{1}{10}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}\right)\)
Nhận xét:
\(\dfrac{1}{9}+\dfrac{1}{10}< \dfrac{1}{8}+\dfrac{1}{8}=\dfrac{1}{4}\)
\(\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{40}+\dfrac{1}{40}=\dfrac{1}{20}\)
\(\Rightarrow S< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)
Vậy \(S=\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{2}\)
10 tháng 7 2018
\(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+...+49+50}\)
\(=\dfrac{1}{\dfrac{2\left(2+1\right)}{2}}+\dfrac{1}{\dfrac{3\left(3+1\right)}{2}}+\dfrac{1}{\dfrac{4\left(4+1\right)}{2}}+...+\dfrac{1}{\dfrac{50\left(50+1\right)}{2}}\)
\(=\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{50.51}\right).2\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{50}-\dfrac{1}{51}\right).2\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{51}\right).2\)
\(=\dfrac{49}{102}.2\)
\(=\dfrac{49}{51}\)
AH
Akai Haruma
Giáo viên
10 tháng 7 2018
Lời giải:
Sử dụng công thức:
\(1+2+....+n=\frac{n(n+1)}{2}\)
\(\Rightarrow \frac{1}{1+2+3+...+n}=\frac{2}{n(n+1)}\)
Do đó:
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+....+\frac{1}{1+2+3+...+49+50}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{50.51}\)
\(\Rightarrow \frac{A}{2}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{50.51}\)
\(\frac{A}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{51-50}{50.51}\)
\(\frac{A}{2}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}=\frac{1}{2}-\frac{1}{51}\)
\(\Rightarrow A=1-\frac{2}{51}=\frac{49}{51}\)
14 tháng 5 2018
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{50}}\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^{49}}\)
\(2A-A=1-\frac{1}{2^{50}}\)
\(A=1-\frac{1}{2^{50}}< 1\)
\(\Rightarrow A< 1\)
DT
Dang Tung
CTVHS
21 tháng 6 2023
\(\dfrac{1}{50}-\dfrac{1}{50.49}-\dfrac{1}{49.48}-...-\dfrac{1}{2.1}\\ =-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{48.49}+\dfrac{1}{49.50}-\dfrac{1}{50}\right)\\ =-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{48}-\dfrac{1}{49}+\dfrac{1}{49}-\dfrac{1}{50}-\dfrac{1}{50}\right)\\ =-\left(1-\dfrac{1}{50}-\dfrac{1}{50}\right)\\ =-\dfrac{24}{25}\)
NT
0
