2L
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
HH
2
26 tháng 2 2020
S = (1/2 + 3/2 + 5/2 + ... + 19/2) + (1+2+3+...+9)
= A + 45
A = \(\frac{1+3+5+...+19}{2}\)
Tử số có số số hạng là: (19-1)/2 + 1 = 10 số
=> A = \(\frac{\left(19+1\right)x10:2}{2}=20x5:2=50\)
=> S = 50 + 45 = 95
LT
26 tháng 2 2020
S= (1+2+3+4+...+9) . \(\left(\frac{1}{2}+\frac{3}{2}+\frac{5}{2}+...+\frac{19}{2}\right)\)
S=(1+9).9:2.\(\frac{1+3+5+7+...+19}{2}\)
S=45.\(\frac{\left[\left(19-1\right):2+1\right].\left(19+1\right):2}{2}\)
S=45.50=2250
S
10 tháng 5 2017
Bài 1:
A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
= \(1-\frac{1}{50}=\frac{49}{50}\)
Bài 2:
Ta có: \(\frac{1}{1^2}=1;\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow A< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1+1-\frac{1}{50}=2-\frac{1}{50}< 2\)
Vậy A < 2
Bài 3:
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
Bài 4:
\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(2S=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)
\(2S-S=\left(6+3+\frac{3}{2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\)
\(S=6-\frac{3}{2^9}=6-\frac{3}{512}=\frac{3069}{512}\)
10 tháng 5 2017
A=1-1/2+1/2-1/3+.............................1/49-1/50
A=1-1/50
A=49/50
20 tháng 3 2018
Ta có :
\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(2S=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)
\(2S-S=\left(6+3+\frac{3}{2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\)
\(S=6-\frac{3}{2^9}\)
\(S=\frac{2^{10}.3-3}{2^9}\)
Vậy \(S=\frac{2^{10}.3-3}{2^9}\)
20 tháng 3 2018
vận dụng 3S lên
xong tìm S nha bn ok
tại k có thời gian nên chỉ giúp thế thôi
DL
1
7 tháng 5 2017
\(S=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2017}\)
\(S=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2017.2018}\)
\(\frac{1}{2}S=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(\frac{1}{2}S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(\frac{1}{2}S=\frac{1}{2}-\frac{1}{2018}\)
\(\frac{1}{2}S=\frac{504}{1009}\)
=> \(S=\frac{1008}{1009}\)
G
3
7 tháng 5 2017
\(S=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2017}\)
\(S=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{2035153}\)
\(S=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{4070306}\)
\(S=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{2017.2018}\)
\(S=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2017.2018}\right)\)
\(S=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)
\(S=2.\left(\frac{1}{2}-\frac{1}{2018}\right)=2.\frac{504}{1009}=\frac{1008}{1009}\)
Vậy \(S=\frac{1008}{1009}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
25 tháng 7 2018
Câu a
\(S=\frac{3-1}{1x3}+\frac{5-3}{3x5}+\frac{7-5}{5x7}+...+\frac{2019-2017}{2017x2019}.\)
\(S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}=1-\frac{1}{2019}=\frac{2018}{2019}\)
Câu b
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^6}+\frac{1}{3^7}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^5}+\frac{1}{3^6}\)
\(2A=3A-A=1-\frac{1}{3^7}\Rightarrow A=\frac{1}{2}-\frac{1}{2.3^7}\)
VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
28 tháng 2 2020
\(S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{100}\)
\(\Rightarrow2S=2+1+\frac{1}{2}+\frac{1}{2^2}...+\frac{1}{99}\)
\(2S-S=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)
\(\Leftrightarrow2S-S=S=2-\frac{1}{2^{100}}=\frac{2^{101}}{2^{100}}-\frac{1}{2^{100}}=\frac{2^{101}-1}{2^{100}}\)
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
=>\(2S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
=>\(2S-S=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)
=>\(S=1-\frac{1}{2^9}=\frac{511}{512}\)
Vậy \(S=\frac{511}{512}\)
Ta có : \(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^9}\)
\(\Rightarrow2S=1+\frac{1}{2}+\frac{1}{2^3}+....+\frac{1}{2^8}\)
\(\Rightarrow2S-S=1-\frac{1}{2^9}\)
\(\Leftrightarrow S=1-\frac{1}{2^9}\)