\(\Leftrightarrow3x-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\right)=\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\right)\)

\(\Leftrightarrow3x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{2.3}+....+\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(\Leftrightarrow3x-\left(1-\frac{1}{100}\right)=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)

\(\Leftrightarrow3x-\frac{99}{100}=\frac{1}{2}\cdot\frac{189}{380}\)

\(\Leftrightarrow3x-\frac{99}{100}=\frac{189}{760}\)

\(\Leftrightarrow3x=\frac{189}{760}+\frac{99}{100}=\frac{4707}{3800}\)

\(\Leftrightarrow x=\frac{1569}{3800}\)

\(\text{Vậy }x=\frac{1569}{3800}\)

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F = 1- 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100

= 1 - 1/100

= 99/100

nhiều quá !!!

f*** you b****

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 x 2 + 2 x 3 + 3 x 4 +........+ 99 x 100

=> 2 + 6 + 12 +........+ 9900

=> 8 + 12 +.....+ 9900

=> 20 +....+ 9900

=> 20 + 20 + 30 +....+ 9900

=> 70 +....+ 9900

=> ( 9900 x 70 ) : 2

=> 693000 : 2

=> 346500

B = 1 x 2 x 3 + 2 x 3 x 4 +......+ 98 x 99 x100

=> ( 1 x 2 x 3) + ( 2 x 3 x 4 ) +....+ ( 98 x 99 x 100 )

= 6 + 24 +.......+ 970200

=> 28 + 120 +...+ 970200

=> ( 148 x 970200 ) : 2

=> 143589600 : 2

=> 71794900

a)

\(A=1.2+2.3+3.4+...+n.\left(n+1\right)\)

\(3A=1.2.3+2.3.3+3.4.3+...+n.\left(n+1\right).3\)

\(3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n.\left(n+1\right).\left[\left(n+2\right)-\left(n-1\right)\right]\)

\(3A=(1.2.3-0.1.2)+\left(2.3.4-1.2.3\right)+\left(3.4.5-2.3.5\right)+...+\left[n.\left(n+1\right).\left(n+2\right)-\left(n-1\right).n.\left(n+1\right)\right]\)\(3A=-0.1.2+n.\left(n+1\right).\left(n+2\right)\)

\(3A=n.\left(n+1\right).\left(n+2\right)\)

\(A=\dfrac{n.\left(n+1\right).\left(n+2\right)}{3}\)

c)

\(B=1.2.3+2.3.4+...+\left(n-1\right).n.\left(n+1\right)\)

\(4B=1.2.3.4+2.3.4.4+3.4.5.4+...+\left(n-1\right).n.\left(n+2\right).4\)

\(4B=1.2.3.4+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+...+\left(n-1\right).n.\left(n+1\right).\left[\left(n+2\right)-\left(n-2\right)\right]\)\(4B=1.2.3.4+\left(2.3.4.5-1.2.3.4\right)+\left(3.4.5.6-2.3.4.5\right)+...+\left[\left(n-1\right).n.\left(n+1\right).\left(n+2\right)-\left(n-1\right).n.\left(n+1\right).\left(n-2\right)\right]\)\(4B=\left(n-1\right).n.\left(n+1\right).\left(n+2\right)\\ B=\dfrac{\left(n-1\right).n.\left(n+1\right).\left(n+2\right)}{4}\)