ta có :

S = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100

S x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3

S x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)

S x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.

S x 3 = 99x100x101

S = 99x100x101 : 3

S = 333300

      C=1x2+2x3+3x4 +......+99x100

<=>3xC=1x2x3+2x3x3+.....+99x100x3

<=>3xC=1x2x3+2x3x(4-1)+.....+99x100x(101-98)

<=>3xC=1x2x3-1x2x3+2x3x4-2x3x4+......-98x99x100+99x100x101

<=>3xC=99x100x101   <=> C=99x100x101 :3    <=> C= 333300

\(C=1.2+2.3+3.4+...+99.100\)

=> 3C = 1.2.3 + 2.3.3 + 3.4.3 + ...+ 99.100.3

3C = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + ...+ 99.100.(101-98)

3C = 1.2.3 + 2.3.4 -1.2.3 + 3.4.5 - 2.3.4 + ...+ 99.100.101-98.99.100

\(3C=\left(1.2.3+2.3.4+3.4.5+...+99.100.101\right)-\left(1.2.3+2.3.4+...+98.99.100\right)\)

\(3C=99.100.101\Rightarrow C=\frac{99.100.101}{3}=333300\)

Dấu . là dấu nhân

3C = 3 x C

C = 1 x 2 + 2 x 3 + 3 x 4 + ....+ 99 x 100

C x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + .. + 99x100x3

C x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)

C x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100

C x 3 = 99 x 100 x101

C = 99 x 100 x 101 : 3

C = 333 300

\(=\frac{1.2}{99.100}\)

\(=\frac{2}{9900}=\frac{1}{4950}\)

\(\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{99.100}\)

\(=2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)

\(=2.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=2.\left(1-\dfrac{1}{100}\right)=2.\dfrac{99}{100}=\dfrac{99}{50}\)

Đặt A = \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

=> A = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

=> A = 1 - \(\dfrac{1}{100}\) = \(\dfrac{99}{100}\)

=> 1 = \(\dfrac{100}{100}\)

=> A < 1

A =

=> A =

=> A = 1 - =

=> 1 =

=> A < 1

\(\Rightarrow A=5\left(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{99x100}\right)\)

\(\Rightarrow A=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(\Rightarrow A=5\left(1-\frac{1}{100}\right)\)

\(\Rightarrow A=\frac{5x99}{100}=\frac{99}{20}\)

\(A=\frac{5}{1}-\frac{5}{2}+\frac{5}{2}-\frac{5}{3}+\frac{5}{3}-\frac{5}{4}+....+\frac{5}{99}-\frac{5}{100}\)

\(A=\frac{5}{1}+\left(-\frac{5}{2}+\frac{5}{2}\right)+\left(-\frac{5}{3}+\frac{5}{3}\right)+\left(-\frac{5}{4}+\frac{5}{4}\right)+...\left(-\frac{5}{99}+\frac{5}{99}\right)+\frac{5}{100}\)

\(A=\frac{5}{1}+0+0+....+0+\frac{5}{100}\)

\(A=\frac{500}{100}+\frac{5}{100}=\frac{205}{100}=\frac{101}{20}\)

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\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{98\times99}+\frac{1}{99\times100}\)

\(=\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+....+\frac{99-98}{98\times99}+\frac{100-99}{99\times100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

1/1 - 1/101 = 100/101

bằng 100/101

\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}.\)

\(=\frac{1}{1}-\frac{1}{101}\)

\(=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)