a) \(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)

\(F=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)

\(3F=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\)

\(3F=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\)

\(3F=\frac{1}{3}-\frac{1}{33}\)

\(F=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\)

\(F=\frac{1}{3}.\frac{1}{3}-\frac{1}{3}.\frac{1}{33}=\frac{1}{9}-\frac{1}{99}=\frac{11}{99}-\frac{1}{99}=\frac{10}{99}\)

b) \(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(A=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)

\(A=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(A=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\left(\frac{7}{70}-\frac{1}{70}\right)=7.\frac{6}{70}\)

\(A=\frac{7.6}{70}=\frac{1.6}{10}=\frac{1.3}{5}=\frac{3}{5}\)

a, \(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)

\(F=\frac{1}{3}\cdot\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{30\cdot33}\right)\)

\(F=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)

\(F=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{33}\right)\)

\(F=\frac{1}{3}-\frac{10}{33}\)

\(F=\frac{10}{99}\)

a) \(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)

\(F=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)

\(F=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)

\(F=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{33}\right)\)

\(F=\frac{1}{3}.\frac{10}{33}\)

\(F=\frac{10}{99}\)

b) \(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(A=7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)\)

\(A=7.\frac{3}{35}\)

\(A=\frac{3}{5}\)

\(\frac{1}{11x12}+\frac{1}{12x13}+...+\frac{1}{98x99}\)= \(\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+.....+\frac{1}{98}-\frac{1}{99}\)

                                                                       = \(\frac{1}{11}-\frac{1}{99}\)

                                                                        =   \(\frac{8}{99}\)

Học tốt !

\(\frac{1}{11\times12}+\frac{1}{12\times13}+\frac{1}{13\times14}+...+\frac{1}{97\times98}+\frac{1}{98\times99}\)

\(=\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\)

\(=\frac{1}{11}-\frac{1}{99}\)

\(=\frac{8}{99}\)

\(=\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+.....+\frac{1}{98}-\frac{1}{99}\)

\(=\frac{1}{11}-\frac{1}{99}\)

\(=\frac{9}{99}-\frac{1}{99}\)

\(=\frac{8}{99}\)

1/10×11 + 1/11×12 + 1/12×13 + ... + 1/999×1000

= 1/10 - 1/11 + 1/11 - 1/12 + 1/12 - 1/13 + ... + 1/999 - 1/1000

= 1/10 - 1/1000

= 100/1000 - 1/1000

= 99/1000

1/10×11 + 1/11×12 + 1/12×13 + ... + 1/999×1000

= 1/10 - 1/11 + 1/11 - 1/12 + 1/12 - 1/13 + ... + 1/999 - 1/1000

= 1/10 - 1/1000

= 100/1000 - 1/1000

= 99/1000

1/10×11 + 1/11×12 + 1/12×13 + ... + 1/999×1000

= 1/10 - 1/11 + 1/11 - 1/12 + 1/12 - 1/13 + ... + 1/999 - 1/1000

= 1/10 - 1/1000

= 100/1000 - 1/1000

= 99/1000

mới đầu mình cũng định làm như lê thành đạt , nhưng x là dấu nhân thì đâu thể làm theo cách đấy .

vd:1/(10x11)x1/(11x12) khác vs 1/(10x11)+1/(11x12)

hay cậu cho đề sai ????

\(\frac{5}{10.11}+\frac{5}{11.12}+\frac{5}{12.13}+....+\frac{5}{49.50}\)

\(=5.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+.....+\frac{1}{49.50}\right)\)

\(=5.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{49}-\frac{1}{50}\right)\)

\(=5.\left(\frac{1}{10}-\frac{1}{50}\right)\)

\(=5.\frac{2}{25}\)

\(=\frac{2}{5}\)

5/10 x 11 + 5/11 x 12 + 5/12 x 13 + .... + 5/49 x 50

= 5/10 - 5/11 + 5/11 - 5/12 + 5/12 - 5/13 + ....... + 5/49 - 5/50

= 5/10 - 5/50

= 2/5

=2/5 nha bn

S = 728

S =  1x2 + 2x3 + 3x4 + ……………… + 11x12 + 12x13
3S=1x2x3 + 2x3x3 + 3x4x3+ ………. + 11x12x3 + 12x13x3

Ta lấy K = 1x2x3 +2x3x4  + 3x4x5 + ……   + 11x12x13 + 12x13x14
 -     3S  = 1x2x3 + 2x3x3 + 3x4x3+ ………  + 11x12x3  + 12x13x3
            ------------------------------------------------------------------------------------
 K – 3S =     0     + 2x3x1  + 3x4x2 + …… .. + 11x12x10 + 12x13x11

K – 3S =   K – 12x13x14
Từ đó suy ra:   3S = 12x13x14
S = 4x13x14 = 728  
 

Cách 2:
S x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + …. + 11x12x(13-10) + 12x13x(14-11)
S x 3 = 1x2x3 + 2x3x4 – 2x3x1 + 3x4x5 – 3x4x2 + …..+ 11x12x13 – 11x12x10 +12x13x14 – 12x13x11
S x 3 = 12 x 13 x14
S = 4 x 13 x 14
S = 728 

ai k minh minh k lai cho

C=7/10x11+7/11x12+7/12x13+.................+7/69x70

C=1x7/10x11+1x7/11x12+...........+1x7/69x70

C=7(1/10x11+1/11x12+1/12x13+....+1/69x70)

C=7(1/10-1/11+1/11-1/12+1/12-1/13+.......+1/69-1/70)

C=7(1/10-1/70)

C=7(7/70-1/70)

C=7x6/70

C=3/5

a) 3 + 8 + 13 + 18 + ... + 2008 + 2013

Áp dụng công thức tính dãy số ta có :

\(3+8+...+2013=\frac{\left[\left(2013-3\right):5+1\right].\left(2013+3\right)}{2}=\frac{403.2016}{2}=403.1008=406224\)

3 + 8 + 13 + 18 +... + 2008 +2013 ( 403 số hạng )

= ( 3 + 2013 ) x 403 : 2

=406224

* nhận xét: đây là dãy số tăng dần đều có khoảng cách là 5

số số hạng của dãy trên là;

(2013 - 3) : 5 +1 = 403

tổng trên là:

(2013 + 1) x 403 : 2 = 405821

đáp số: 405821

NHA