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1990x1990-1992×1988
=1990×(1988 + 2)-(1990 + 2)×1988
=(1990×1988+1990×2)-(1990×1988+2×1988)
=1990×1988+1990×2-1990×1988-2×1988
=1990×2-2×1988
=2×(1990 - 1988)
=2×2
=4
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
A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
A x 3 = 99x100x101
A = 99x100x101 : 3
A = 333300
Ta có:
\(A=1.2+2.3+3.4+...+99.100\)
\(\Rightarrow3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)
\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)
\(\Leftrightarrow3A=99.100.101\Leftrightarrow A=\frac{99.100.101}{3}=333300\)
\(B=1.2.3+2.3.4+4.5.6+...+98.99.100\)
\(\Rightarrow4B=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+4.5.6.\left(7-3\right)+...+98.99.100.\left(101-97\right)\)
\(\Rightarrow4B=1.2.3.4+2.3.4.5-1.2.3.4+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100\)
\(\Leftrightarrow4B=98.99.100.101\Leftrightarrow B=\frac{98.99.100.101}{4}=24497550\)
\(A=1.2+2.3+3,4+...+1999.2000\)
\(=>3A=1.2.3+2.3.3+3.4.3+...+1999.2000.3\)
\(=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+.....+1999.2000.\left(2001-1998\right)\)
\(=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+1999.2000.2001-1998.1999.2000\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+1999.2000.2001-1998.1999.2000\)
\(=1999.2000.2001\)
\(=>A=\frac{1999.2000.2001}{3}=......\) (bn dùng máy tính)
b,xem lại chỗ 3.2
c,tính 4C , biến đổi tương tự câu a
