A=2(1-3)+4(5-3)+ 6(5-7)+...+50(49-57)

A=-4-8-12-...-100 = -(4+8+12+...+100) (tính tổng cấp số cộng)

Ta có:A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+......+\dfrac{1}{49.50}+\dfrac{1}{50.51}\)

A=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.......+\dfrac{1}{49}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{51}\)

A=1-\(\dfrac{1}{51}=\dfrac{50}{51}\)

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}+\dfrac{1}{50.51}\)

\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{51}\)

\(A=\dfrac{1}{1}-\dfrac{1}{51}\)

\(A=\dfrac{50}{51}\)

A = \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}+\dfrac{1}{50.51}\)

= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{51}\)

= \(1-\dfrac{1}{51}\)

= \(\dfrac{50}{51}\)

=43.61

Lời giải:

$A=1.2+2.3+3.4+...+50.51$
$3A=1.2(3-0)+2.3(4-1)+3.4(5-2)+...+50.51(52-49)$

$=(1.2.3+2.3.4+3.4.5+...+50.51.52)-(0.1.2+1.2.3+2.3.4+....+49.50.51)$

$=50.51.52$

$\Rightarrow A=50.51.52:3=44200$

Ta có: 3S = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2)  + .....+ 50.51.(52 -49) 

              = 1.2.3 - 0  + 2.3.4 - 1.2.3 + 3.4.5 -2.3.4 + .....+ 50.51.52 - 49.50.51

 3S = 50.51.52

  S = 50.17.52 =44200

3S = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2)  + .....+ 50.51.(52 -49) 

    = 1.2.3 - 0  + 2.3.4 - 1.2.3 + 3.4.5 -2.3.4 + .....+ 50.51.52 - 49.50.51

 3S = 50.51.52

S = 50.17.52 =44200

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}=\frac{49}{50}\)

\(B=1.2+2.3+3.4+...+49.50\)

\(3B=1.2.3+2.3.3+3.4.3+...+49.50.3\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)

\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)

\(=49.50.51\)

\(B=\frac{49.50.51}{3}=49.50.17\)

\(50^2.A-\frac{B}{17}=49.50-49.50=0\)

C = 1.2 + 2.3 + 3.4 + ... + 99.100

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

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

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

       = 99.100.101

=>C = 99.100.101 : 3 = 33.100.101 =     333300

Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 32.33

=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 32.33.34

=> 3S = 32.33.34

=> S = \(\frac{32.33.34}{3}=11968\)