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Tính tổng: 1x2 + 2x3 + 3x4 + 4x5 +.............+ 99x100
Gọi biểu thức trên là A, ta có :
A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
3A= 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
3A = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
3A = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
3A = 99x100x101
A = 99x100x101 : 3
A = 333300
Ta có:
\(D=1.2+2.3+3.4+4.5+...+99.100\)
\(\Leftrightarrow3D=1.2.\left(3-0\right)+2.3+\left(4-1\right)+3.4+\left(5-2\right)+4.5.\left(6-3\right)+...+99.100.\left(101-98\right)\)
\(\Leftrightarrow3D=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100\)
\(\Leftrightarrow3D=99.100.101\Leftrightarrow D=\frac{99.100.101}{3}=333300\)
\(B=1.3+2.4+3.5+4.6+...+99.101\)
\(\Leftrightarrow B=\left(1.3+3.5+...+99.101\right)+\left(2.4+4.6+...+98.100\right)\)
\(\Leftrightarrow6B=\left(1.3.\left(5-\left(-1\right)\right)+3.5.\left(7-1\right)+...+99.101.\left(103-97\right)\right)+\left(2.4.\left(6-0\right)+4.6.\left(8-2\right)+...+98.100.\left(102-96\right)\right)\)
\(\Leftrightarrow B=\frac{99.101.103+3}{6}+\frac{98.100.102}{6}=338250\)
Vì các bước gần tương tự như bài a) nên mình bỏ bước.
\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\)
\(\Leftrightarrow C=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)
\(\Leftrightarrow C=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(\Leftrightarrow C=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{49.50}\right)=\frac{1}{2}.\frac{612}{1225}=\frac{306}{1225}\)
a, 200 - 3( x - 16 ) = 20
3( x - 16 ) = 200 - 20 = 180
x - 16 = 180 : 3 = 60
x = 60 + 16 = 76
b, 5 + 10 + 15 + .............. + 95 + 100 + 105 = 1200
c, x + ( 99 - 97 + 95 - 93 + ............ + 7 - 5 + 3 - 1 ) = 100
x + ( 2 . 25 ) = 100
x + 50 = 100
x = 100 - 50 = 50
****, thks
\(3\times\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)
\(=3\times\left(\dfrac{50}{100}-\dfrac{1}{100}\right)\)
\(=3\times\dfrac{49}{100}=\dfrac{147}{100}\)
3∙(\(\dfrac{1}{2}\)-\(\dfrac{1}{100}\))
=3∙(\(\dfrac{50}{100}\)-\(\dfrac{1}{100}\))
=3∙ \(\dfrac{49}{100}\)
= \(\dfrac{147}{100}\)