Question

Cho dãy số: 31;32;33;...;n.Tìm n biết: 31+32+33+...+n=1950

 

Answer 1

\(31+32+33+...+n=1950\)

\(\Leftrightarrow\dfrac{\left(n+31\right)\left(n-30\right)}{2}=1950\)

\(\Leftrightarrow n^2+n-2970=0\)

\(\Leftrightarrow n=54\)

Answer 2

\(31+32+33+...+n=1950\\ \Rightarrow\left(n+31\right)\left(n-31+1\right):2=1950\\ \Rightarrow\left(n+31\right)\left(n-30\right)=3900\)

Ta thấy \(\left(n+31\right)\) cách \(\left(n-30\right)\) 60 đơn vị nên phân tích \(3600=39\times100\)

\(\Rightarrow\left\{{}\begin{matrix}n+31=100\\n-30=39\end{matrix}\right.\Rightarrow n=69\)