Question

Tính giá trị biểu thức:

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

Giúp mk nhe!!!! Mai mình thi rùi!!!!!!!

Answer 1

Ta có: \(1+2+3+...+n=\frac{n.\left(n+1\right)}{2}\)

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

\(Q=\frac{1}{\frac{2.\left(2+1\right)}{2}}+\frac{1}{\frac{3.\left(3+1\right)}{2}}+....+\frac{1}{\frac{10.\left(10+1\right)}{2}}\)

\(Q=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+....+\frac{1}{\frac{10.11}{2}}\)

\(Q=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{10.11}\)

\(\frac{1}{2}Q=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{10.11}\)

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

=>\(Q=\frac{9}{22}.2=\frac{9}{11}\)

Answer 2

 

\(Q=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{55}\\ \Rightarrow\frac{1}{2}Q=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{110}\)

Tiếp theo tự tính nhé