Question

thực hiện phép tính

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

Answer 1

Cần lắm ko

Answer 2

Đỗ Ngọc Hải cần gấp

Answer 3

 Đúng 0             

Answer 4

\(=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+...+\frac{1}{\frac{50.51}{2}}=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{50.51}\right)\)
\(=2.\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{51-50}{50.51}\right)=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{51}\right)=1-\frac{2}{51}=\frac{49}{51}\)

Answer 5

quên cách làm rồi

Answer 6

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

Ta có. \(\frac{2}{2.\left(1+2\right)}+\frac{2}{2.\left(1+2+3\right)}+...+\frac{2}{2.\left(1+2+...+50\right)}\)

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

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

= 2.( \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{50.51}\))

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

= 2. ( \(\frac{1}{2}-\frac{1}{51}\))

= 2. ( \(\frac{51}{102}-\frac{2}{102}\))

= 2. \(\frac{49}{102}\)

= \(\frac{49}{51}\)

Answer 7

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

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

\(=1:\frac{3.2}{2}+1:\frac{4.3}{2}+1:\frac{5.4}{2}+...+1:\frac{51.50}{2}\)

\(=1.\frac{2}{3.2}+1.\frac{2}{4.3}+1.\frac{2}{5.4}+...+1.\frac{2}{51.50}\)

\(=\frac{2}{3.2}+\frac{2}{4.3}+\frac{2}{5.4}+...+\frac{2}{51.50}\)

\(=2.\left(\frac{1}{3.2}+\frac{1}{4.3}+\frac{1}{5.4}+...+\frac{1}{51.50}\right)\)

\(=2.\left(\frac{3-2}{3.2}+\frac{4-3}{4.3}+\frac{5-4}{5.4}+...+\frac{51-50}{51.50}\right)\)

\(=2.\left(\frac{3}{2.3}-\frac{2}{2.3}+\frac{4}{4.3}-\frac{3}{4.3}+\frac{5}{5.4}-\frac{4}{5.4}+...+\frac{51}{51.50}-\frac{50}{50.51}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{51}\right)\)

\(=2.\left(\frac{51}{102}-\frac{2}{102}\right)\)

\(=2.\frac{49}{102}\)

\(=\frac{49}{51}\)

Answer 8

49/51