Question

Tìm n thuộc N thõa : \(1+\frac{1}{3}+\frac{1}{6}+\frac{2}{n\left(n+1\right):2}=1\frac{1997}{1998}\)

 

Answer 1

Lời giải:
$1+\frac{1}{3}+\frac{1}{6}+....+\frac{1}{n(n+1):2}=1\frac{1997}{1998}$

$1+\frac{2}{6}+\frac{2}{12}+....+\frac{2}{n(n+1)}=1\frac{1997}{1998}$

$1+2[\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{n(n+1)}]=1\frac{1997}{1998}$
$1+2(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{n}-\frac{1}{n+1})=1\frac{1997}{1998}$

$2(\frac{1}{2}-\frac{1}{n+1})=\frac{1997}{1998}$

$1-\frac{2}{n+1}=\frac{1997}{1998}$

$\frac{2}{n+1}=1-\frac{1997}{1998}=\frac{1}{1998}$

$n+1=1998.2=3996$
$n=3995$