\(1+2+3+...+n=4753\)
\(\Rightarrow\frac{\left(n+1\right)\left[\left(n-1\right):1+1\right]}{2}=4753\)
\(\Rightarrow\frac{n\left(n+1\right)}{2}=4753\)
\(\Rightarrow n\left(n+1\right)=4753.2=9506=97.98\)
\(\Rightarrow n=97\)
1+2+3+4+...+n = 4753
(1+n).n:2 = 4753
(1+n).n = 9506 = 98.97 ( do 1+n và n là 2 số tự nhiên liên tiếp)
=> n = 97
\(1+2+3+4+...+n=4753\)
\(\Leftrightarrow\frac{n\left(n+1\right)}{2}=4753\)
\(\Leftrightarrow n\left(n+1\right)=4753\times2\)
\(\Leftrightarrow n\left(n+1\right)=9506\)
\(\Leftrightarrow97\times98=9506\)
\(\Rightarrow n=97\)
