Question

[Chinh Phạm Thị]

1/1+2+1/1+2+3+1/1+2+3+4+...+1/1+2+3+...+n=0.92

Answer 1

[Xyz OLM]

\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+n}=0,92\)

\(\Leftrightarrow\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+...+\frac{1}{\frac{n\left(n+1\right)}{2}}=\frac{23}{25}\)

<=> \(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{n\left(n+1\right)}=\frac{23}{25}\)

<=> \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}=\frac{23}{50}\)

<=>  \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{23}{50}\)

=> \(\frac{1}{2}-\frac{1}{n+1}=\frac{23}{50}\)

=> \(\frac{1}{n+1}=\frac{1}{25}\)

=> n + 1 = 25 

=> n = 24

Vậy n = 24