Đặt S = 1 x 2 + 2 x 3 + ..... + 9 x 10

3S = 1 x 2 x (3-0) + 2 x 3 x (4 - 1) + ....... + 9 x 10 x (11 - 8)

3S = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + ......... + 9 x 10 x 11 - 8 x 9 x 10

3S = ( 1 x 2 x 3 -1 x 2 x 3) + ( 2 x 3 x4 - 2 x 3 x 4) + ..... + (8 x 9  x 10 - 8 x 9 x 10) + 9 x 10 x 11

3S = 9 x 10 x 11 = 990

S = 990 : 3 = 330     

A=1/1*2+1/2*3+...+1/9*10

=1-1/2+1/2-1/3+...+1/9-1/10

=(1-1/10)+(1/2-1/2)+...+(1/9-1/9)

=(10/10-1/10)+0+...+0=9/10

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}=\frac{9}{10}\)

1/10x(2/9x9/2)x(3/8x8/3)x(5/6x6/5)=1/10x1x1x1=1/10

tổng quát: \(\frac{1}{n.\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

Ta có: \(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)

\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{3}-\frac{1}{10}=\frac{10}{30}-\frac{3}{30}=\frac{7}{30}\)

 \(\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\) \(\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}\)

\(=\dfrac{1+2+3+4+5+6+7+8+9}{10}\)

\(=\dfrac{45}{10}=\dfrac{9}{2}\)

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9/10 = 44/10 = 22/5

bằng 9/2 nhé

1 + 9 + 2 + 8 + 3 + 7 + 4 + 6 + 5 + 5 + 6 + 4 + 7 + 3 + 8 + 2 + 9 + 1

= 10 + 10 + 10 +10 +10 + 10 + 10 + 10 + 10

= 10 x 9

= 90

1+9+2+8+3+7+4+6+5+5+6+4+7+3+8+2+9+1

=10+10+10+10+10

=50

Số số hạng là :

(50-1):1+1=50(số)

Tổng dãy số là

(50+1)x50:2=1275

Đ/S:1275

k đúng cho m nha

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 +...+ 50

= 1 + 50 + 2 + 49 + 3 + 48 + 4 + 47 + ... + 25 + 26

= (1 + 50) + (2 + 49) + (3 + 48) + (4 + 47) + ... + (25 + 26)

= 51 + 51 + 51 + 51 + ... + 51 (25 số hạng 51 vì 50 : 2 = 25)

= 51 x 25

= 1275

Dãy trên có số số hạng là :

( 50 - 1 ) : 1 + 1 = 50 ( số )

Tổng là :

( 50 + 1 ) x 50 : 2 = 1275

Đáp số : 1275

\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+..+10}\)

\(=\frac{1}{3}+\frac{1}{6}+...+\frac{1}{55}\)

\(=2\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=2.\left(\frac{1}{2}-\frac{1}{11}\right)=2.\frac{9}{22}=\frac{9}{11}\)