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B = \(1+\frac{1}{3}+\frac{1}{6}+....+\frac{1}{630}=1+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{1260}\)
B = \(1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{35.36}\right)\)
B = \(1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{35}-\frac{1}{36}\right)\)
B = \(1+2\left(\frac{1}{2}-\frac{1}{36}\right)=1+2.\frac{17}{36}\)
B = \(1+\frac{17}{18}\)
B = \(\frac{35}{18}\)
\(A=\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{99x101}\)
\(A\)\(x2=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{99x101}\)
\(A\)\(x2=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(A\)\(x2=1-\frac{1}{101}=\frac{100}{101}\)
\(A=\frac{100}{101}:2=\frac{100}{101}x\frac{1}{2}=\frac{50}{101}\)
Ta có:
A = 1 + 3 + 5 + 7 +... + 101
A = \(\frac{102.51}{2}=2601\)
M = 16 - 18 + 20 - 22 + 24 - 26 + .. + 64 - 66 + 68
M = ( 16 - 18 ) + ( 20 - 22 ) + ( 24 - 26 ) + ... + ( 64 - 66 ) + 68
M = (- 2 + - 2 + -2 + ... + - 2 ) + 68
M = 25/2 . ( - 2 ) + 68
M = -25 + 68
M = 43
H = ( 1 + 2 + 3 +...+ 99 ) x ( 13 x 15 - 12 x 15 - 15 )
H = ( 1 + 2 + 3 +...+ 99 ) x { (13 - 12 - 1) x 15 }
H = ( 1 + 2 + 3 +...+ 99 ) x 0
H = 0
G = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + 10 - 11 - 12 + 13 + 14 - ... + 301 + 302
G = ( 1 + 2 ) + ( -3 - 4 ) + ( 5 + 6 ) + ( -7 - 8 ) + ( 9 + 10 ) + ( - 11 - 12 ) + ( 13 + 14 ) -...+ ( 301 + 302 )
G = ( 3 - 7 ) + ( 11 - 15 ) + ( 19 - 23 ) + 27 - ... + 603
G = -4 + - 4 + -4 + 27 - ... + 603
G = 75 x ( -4 ) + 603
G = -300 + 603
G = 303
2.
a) 1 + 2 + 3 + 4 +...+ 99 + 100 + 2 x X = 5052
= > \(\frac{100.101}{2}\)+ 2 x X = 5052
= > 5050 + 2 x X = 5052
= > 2X = 2
= > X = 1
A=1/1*2+1/2*3+...+1/9*10
=1-1/2+1/2-1/3+...+1/9-1/10
=(1-1/10)+(1/2-1/2)+...+(1/9-1/9)
=(10/10-1/10)+0+...+0=9/10
1/3*5+1/5*7+1/7*9+...+1/97*99
=1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99
=1/3-1/99
32/99
Đặt : \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{99\cdot101}\)
\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}\)
\(2A=\frac{100}{101}\)
\(A=\frac{100}{101}\cdot\frac{1}{2}=\frac{50}{101}\)
Ta có:
a)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)
b)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{210}\)
\(=2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}\right)\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{21}\right)=2.\frac{19}{42}=\frac{19}{21}\)
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+..+10}\)
\(=\frac{1}{3}+\frac{1}{6}+...+\frac{1}{55}\)
\(=2\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=2.\left(\frac{1}{2}-\frac{1}{11}\right)=2.\frac{9}{22}=\frac{9}{11}\)
1/2*3+1/3*4+1/4*5+1/5*6+1/6*7+...+1/98*99+1/99*100
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{50}{100}-\frac{1}{100}\)
\(=\frac{49}{100}\)
1/2 x 3 + 1/3 x 4 + 1/4 x 5 + 1/5 x 6 + 1/6 x 7 + ....... + 1/98 x 99 + 1/99 x 100
= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ..... + 1/98 - 1/99 + 1/99 - 1/100
= 1/2 - 1/100
= 49/100
A = \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\)
A = \(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
A = \(\frac{1}{2}\left(1-\frac{1}{101}\right)=\frac{1}{2}.\frac{100}{101}\)
A = \(\frac{50}{101}\)
B = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{630}\)
B = \(1+\frac{2}{6}+\frac{2}{12}+\frac{1}{20}+...+\frac{2}{1260}\)
B = \(1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{35.36}\right)\)
B = \(1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{35}-\frac{1}{36}\right)\)
B = \(1+2\left(\frac{1}{2}-\frac{1}{36}\right)=1+2.\frac{17}{36}\)
B = \(1+\frac{17}{18}\)
B = \(\frac{35}{18}\)
Quá dễ