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

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

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

\(=2\left(\frac{1}{2.3}+\frac{1}{3\cdot4}+.....+\frac{1}{10\cdot11}\right)\)

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

\(=2\left(\frac{1}{2}-\frac{1}{11}\right)\)

\(=2\cdot\frac{9}{22}\)

\(=\frac{9}{11}\)

Đặt \(B=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{210}\)

  \(\frac{1}{2}B=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}\)

  \(\frac{1}{2}B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\)

   \(\frac{1}{2}B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\)

   \(\frac{1}{2}B=\frac{1}{2}-\frac{1}{21}\)

 \(\Rightarrow B=\frac{\frac{1}{2}-\frac{1}{21}}{\frac{1}{2}}=\frac{19}{21}\)

\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+50}\)

\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{\frac{\left(1+50\right).50}{2}}\)

\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{1275}\)

\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{2550}\)

\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+..+\frac{2}{50.51}\)

\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{51}\right)=2\cdot\frac{49}{102}=\frac{49}{51}\)

cộng hết tất cả 1/1+2+3+.....+10 thì ta chỉ cần cộng 1+2+3+4+5+6+7+8+9+10 là xong rồi tự tính

\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.............+\frac{1}{1+2+3+......+10}\)

= \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+..............+\frac{1}{45}\)

Đến đây bạn làm tiếp nhé

1 ) 

m = 3 

n = 2 

biết vậy nhưng ko biết cách giải

có : Q = [ 2 + 2^2 ] + [ 2^3 +2^4] + ... + [2^9 +  2^10]

Q = 2 [1+2] +2^3[1 +2]+ ...+ 2^9 [1+2]

Q = 2 . 3+2^3 .3 +... + 2^9 .3

Q = 3. [ 2 + 2^3 +... + 2^9]

Vậy Q chia hết cho 3

A = \(\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+...+\frac{1}{9}.\frac{1}{10}\)

A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

A = \(1-\frac{1}{10}\)

A = \(\frac{9}{10}\)

1/2=1-1/2 ; 1/2.1/3=1/2-1/3 ; 1/3.1/4=1/3-1/4...v...v

Vậy A bằng: 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5.............+1/8-1/9+1/9-1/10

                =1-1/10=9/10