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\(E=1\cdot1+2\cdot2+3\cdot3+...+15\cdot15\)
\(=1\cdot\left(2-1\right)+2\cdot\left(3-1\right)+3\cdot\left(4-1\right)+...+15\cdot\left(16-1\right)\)
\(=1\cdot2-1+2\cdot3-2+3\cdot4-3+...+15\cdot16-15\)
\(=\left(1\cdot2+2\cdot3+3\cdot4+...+15\cdot16\right)-\left(1+2+3+...+15\right)\)
\(=1360-120=1240\)
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Ta có : B = 1.1 + 2.2 + 3.3 + ... + 1999.1999
= 1.(2 - 1) + 2.(3 - 1) + 3(4 - 1) + ... + 1999(2000 - 1)
= (1.2 + 2.3 + 3.4 + ... + 1999.2000) - (1 + 2 + 3 + .... + 1999)
Đặt A = 1.2 + 2.3 + 3.4 + ... + 1999.2000
=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 1999.2000.3
= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + .... + 1999.2000.(2001 - 1998)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 1999.2000.2001 - 1998.1999.2000
= 1999.2000.2001
=> A = 1999.2000.2001/3
Khi đó B = A - (1 + 2 + 3 + .... + 1999)
= 1999.2000.2001/3 - 1999.(1999 + 1)/2
= 1999.2000.667 - 1999.1000
= 1999.(2000.667 - 1000)
= 1999 . 1 333 000
Vậy B = 1999 . 1333000
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S5=5x5-(4x4-(3x3-(2x2-1x1)))
S2011=2001x2001-(2000x2000-(1999x1999-(....)))
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A=1x2+2x3+3x4+...+49x50
3A= 3(1.2+2.3+3.4+...+49.50)
3A= 1.2.3+2.3.3+3.4.3+...+49.50.3
3A= 1.2.(3-0)+2.3(4-1)+3.4(5-2)+...+49.50.(51-48)
3A= 0.1.2-1.2.3+1.2.3-2.3.4+2.3.4-3.4.5+...+48.49.50-49.50.51
3A= 49.50.51
A= 49.50.51/3=41650
B=1x3+3x5+5x7+...+99x101
B=1/1.3 +1/3.5 +...+1/99.101
2B=2/1.3 + 2/3.5 +...+2/99.101
2B=1-1/3+1/3-1/5+...+1/99-1/101
2B=1-1/101
2B=100/101
B=100/101:2=100/202
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- \(B=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{93.97}\)
\(4.B=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{93.97}\)
\(4.B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\)
\(4.B=1-\frac{1}{97}\)
\(4.B=\frac{96}{97}\)
\(B=\frac{96}{97}:4\)
\(B=\frac{24}{97}\)
E = 1 x 1 + 2 x2 + 3 x 3 + . . . + 15 x 15
E = 1( 2 -1 ) + 2 ( 3 -1 ) + 3 ( 4-1 ) + . . . + 15 ( 16 -1 )
E = 1.2 -1 + 2.3 - 2 + 3.4 - 3 + . . . + 15 . 16 - 15
E = ( 1.2 + 2.3 + 3.4+ . . . + 15 .16 ) - ( 1 + 2 + 3 + . . . + 15 )
E = 1360- 120
E = 1240
Tk mk nha