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\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{49.50}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+...+\left(\frac{1}{49}-\frac{1}{49}\right)-\frac{1}{50}\)
\(=\frac{1}{2}-\frac{1}{50}=\frac{12}{25}\)
~ Hok tốt ~
\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{49.50}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{50}\right)=2.\frac{12}{25}=\frac{24}{25}\)
1/1x2+1/2x3+...+1/49x50
=1-1/2+1/2-1/3+.....+1/49-1/50
=1-1/50(1)
Ta co 1(2)
So sanh (1) voi (2) ta thay 1-1/50<1
=>1/1x2+...+1/49x50<1
(Phuong phap khu)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)
=\(\frac{1}{1}-\frac{1}{50}=\frac{50}{50}-\frac{1}{50}=\frac{49}{50}<1\)
Vậy \(\frac{49}{50}<1\)
đề có sai không đó bạn làm gì tích 2 số tự nhiên liên mà = 900 chớ
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
a) \(\frac{3}{40}+\frac{5}{3}+\frac{7}{60}=\frac{9}{120}+\frac{200}{120}+\frac{14}{120}=\frac{223}{120}\)
b) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{19.20}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}=1-\frac{1}{20}=\frac{19}{20}\)
H = \(\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{2.3}-\frac{1}{2.3.4}+\frac{1}{3.4}-\frac{1}{3.4.5}+...+\frac{1}{99.100}-\frac{1}{99.100.101}\)
\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{99.100.101}\right)\)
Đặt G = \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
= \(1-\frac{1}{100}\)
= \(\frac{99}{100}\)
Đặt K = \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{99.100.101}\right)\)
=>2K = \(\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{99.100.101}\right)\)
= \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\)
= \(\frac{1}{1.2}-\frac{1}{100.101}\)
= \(\frac{1}{2}-\frac{1}{10100}\)
= \(\frac{5049}{10100}\)
=> K =\(\frac{5049}{10100}:2=\frac{5049}{10100}.\frac{1}{2}=\frac{5049}{20200}\)
Thay G,K vào H ta có :
H = \(\frac{99}{100}-\frac{5049}{20200}\)
Tự tính :)
\(H=\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{2.3}-\frac{1}{2.3.4}+...+\frac{1}{99.100}-\frac{1}{99.100.101}\)
\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.34}+...+\frac{1}{99.100.101}\right)\)
\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{99.100.101}\right)\)
\(=\left(1-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(=\frac{99}{100}-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)=\frac{99}{100}-\frac{1}{2}.\frac{5049}{10100}=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)
=1/2-1/3+1/3-1/4+.....+1/49+1/50
=1/2-1/50
=25/50-1/50
=24/50
=12/25
\(\frac{1}{2x3}\)+ \(\frac{1}{3x4}\)+ ... + \(\frac{1}{49x50}\)
= \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ ... + \(\frac{1}{49}\)- \(\frac{1}{50}\)
= \(\frac{1}{2}\)- \(\frac{1}{50}\)
=\(\frac{12}{25}\)