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BV
0
TN
1
13 tháng 5 2018
Ta có :
\(A=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+...+\frac{1}{97.99}+\frac{1}{98.100}\)
\(A=\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{4}-\frac{1}{6}\right)+...+\frac{1}{2}.\left(\frac{1}{97}-\frac{1}{99}\right)+\frac{1}{2}.\left(\frac{1}{98}-\frac{1}{100}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{98}-\frac{1}{100}\right)\)
\(A=\frac{1}{2}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{97}+\frac{1}{98}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}-...-\frac{1}{99}-\frac{1}{100}\right)\)
\(A=\frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{99}-\frac{1}{100}\right)< \frac{1}{2}.\left(1+\frac{1}{2}\right)=\frac{3}{4}\)
2 tháng 7 2018
a) Số số hạng của dãy A là: (2020-5):2+1 = 404 (số)
Tổng A là: (2020+5)x404:2=409050
b) \(B=\frac{2}{1\times3}+\frac{2}{3\times5}+....+\frac{2}{99\times101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
c) \(C=\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+...+\frac{1}{98\times100}\)
\(=\frac{1}{2}\times\left(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+....+\frac{2}{98\times100}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{100}\right)=\frac{1}{2}\times\frac{99}{100}=\frac{99}{200}\)
Vậy .....
N
2 tháng 7 2018
A = 5 + 10 + 15 + ... + 2015 + 2020
Số số hạng là : 404
A = 409050
\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(B=1-\frac{1}{101}=\frac{101-1}{101}=\frac{100}{101}\)
\(C=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+...+\frac{1}{98\cdot100}\)
\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{2}\cdot\left(\frac{1}{4}-\frac{1}{6}\right)+\frac{1}{2}\cdot\left(\frac{1}{6}-\frac{1}{8}\right)+...+\frac{1}{2}\cdot\left(\frac{1}{98}-\frac{1}{100}\right)\)
\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{2}\cdot\frac{49}{100}=\frac{49}{200}\)
NQ
2
6 tháng 8 2016
\(S=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{97.99}+\frac{1}{98.100}\)
\(S< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{97.98}+\frac{1}{98.99}\)
\(S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\)
\(S< 1-\frac{1}{99}< 1\)
=> S < 1
26 tháng 3 2017
a) Đặt \(A=\frac{1^2}{1.2}+\frac{2^2}{2.3}+.........+\frac{100^2}{100.101}\)
\(\Rightarrow A=\left(1^2+2^2+..........+100^2\right)\)\(.\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{100.101}\right)\)
\(\Rightarrow A=\left(1^2+2^2+......+100^2\right).\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A=\left(1^2+2^2+......+100^2\right).\left(1-\frac{1}{101}\right)\)
\(\Rightarrow A=\left(1^2+2^2+.....+100^2\right).\left(\frac{100}{101}\right)\)(a)
Đặt \(M=\left(1^2+2^2+........+100^2\right)\)
\(\Rightarrow M=1.1+2.2+.....+100.100\)
\(\Rightarrow M=1.\left(2-1\right)+2.\left(3-1\right)+....+100.\left(101-1\right)\)
\(\Rightarrow M=\left(1.2-1\right)+\left(2.3-2\right)+.....+\left(100.101-100\right)\)
\(\Rightarrow M=\left(1.2+2.3+.....+100.101\right)-\left(1+2+......+100\right)\)
\(\Rightarrow M=\left(1.2+2.3+......+100.101\right)-5050\)(1)
Đặt \(N=1.2+2.3+....+100.101\)
\(\Rightarrow3.N=1.2.3+2.3.3+......+100.101.3\)
\(\Rightarrow3N=1.2.\left(3-0\right)+2.3.\left(4-1\right)+......+100.101.\left(102-99\right)\)
\(\Rightarrow3N=\left(1.2.3-0\right)+\left(1.2.3-2.3.4\right)+.......+\left(100.101.102-100.101.99\right)\)
\(\Rightarrow3N=100.101.102-0\)
\(\Rightarrow N=343400\)
Thay N = 343400 vào 1) ta được:
M = 343400 - 5050
=> M = 338350
Thay M = 338350 Vào (a) ta được:
A = 338350 . \(\frac{100}{101}\)
=> \(A=\frac{33835000}{101}\)
Vậy \(\frac{1^2}{1.2}+\frac{2^2}{2.3}+.........+\frac{100^2}{100.101}=\frac{33835000}{101}=335000\)
b) Đặt \(B=\frac{2^2}{1.3}+\frac{3^2}{2.4}+..........+\frac{59^2}{58.60}\)
\(\Rightarrow B=\left(2^2+3^2+........+59^2\right).\left(\frac{1}{1.3}+\frac{1}{2.4}+.....+\frac{1}{58.60}\right)\)
Đặt \(G=2^2+3^2+.........+59^2\)VÀ \(H=\frac{1}{1.3}+\frac{1}{2.4}+.........+\frac{1}{58.60}\)
\(\Rightarrow G=2.2+3.3+.......+59.59\) VÀ \(2.H=\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{58.60}\)
Rồi bạn làm như ở phần a) ý
22 tháng 4 2017
\(A=1.3+2.4+3.5+.............+97.99+98.100\)
\(A=\left(2-1\right)\left(2+1\right)+\left(3-1\right)\left(3+1\right)+.............+\left(99-1\right)\left(99+1\right)\)
\(A=2^2-1+3^2-1+..............+99^2-1\)
\(A=1+2^2+3^2+............+99^2-99\)
Mà :
\(1+2+2^2+...........+n^2=\dfrac{\left(n+1\right)\left(n+2\right)}{6}\)
\(\Rightarrow A=\dfrac{99\left(99+1\right)\left(99+2\right)}{6}-99=\dfrac{99.100.101}{6}-99\)
\(A=166650-99=166551\)
~ Học tốt ~
\(A=\frac{2^2}{1.3}+\frac{3^2}{2.4}+\frac{4^2}{3.5}+...+\frac{99^2}{98.100}\)
\(A=\frac{2.2}{1.3}+\frac{3.3}{2.4}+\frac{4.4}{3.5}+...+\frac{99.99}{98.100}\)
\(A=\frac{2}{1}+\frac{99}{100}\)
\(A=\frac{200}{100}+\frac{99}{100}=\frac{299}{100}\)
Hok tốt