Question

Tìm n :

1/2 + 1/6 + 1/12 + .... + 1/n = 39/40

Giúp với

Answer 1

n=1560

Answer 2

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{n}=\frac{39}{40}\)

Đặt \(n=x\left(x+1\right)\);ta được : 

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}=\frac{39}{40}\)

\(\Leftrightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{39}{40}\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{39}{40}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{39}{40}\)

\(\Leftrightarrow\frac{1}{x+1}=1-\frac{39}{40}=\frac{1}{40}\)

\(\text{Vậy }:x+1=40\Rightarrow x=39\)

\(\Rightarrow n=39.\left(39+1\right)=39.40=1560\)

Answer 3

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{n}=\frac{39}{40}\)

\(\Leftrightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}=\frac{39}{40}\)

\(\Leftrightarrow1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...-\frac{1}{n-1}+\frac{1}{n}=\frac{39}{40}\)

\(\Leftrightarrow1+\frac{1}{n}=\frac{39}{40}\)

\(\Leftrightarrow\frac{1}{n}=\frac{1}{40}\)

\(\Leftrightarrow n=40\)