Tổng A có (x-1):4+1 số hạng
Nếu nhóm số đầu vs só cuối ta được \(\frac{\left(x-1\right):4+1}{2}\)ngoặc.
\(\Rightarrow\)\(\frac{\left(x+1\right)\left(x-1\right):4+1}{2}\)=501502
\(\Rightarrow x:4+1=501502.2=1003004\)
\(\Rightarrow x:4=1003004-1=1003003\)
\(\Rightarrow x=4\)
Vậy x=4
Ta có:
\(5=2+3;9=4+5;13=6+7;17=8+9;...\)
Do vậy \(x=a+\left(a+1\right)\) \(\left(x\in N\right)\)
Nên \(1+5+9+13+17+...+x=1+2+3+4+5+6+7+8+9+...+\left[a+\left(a+1\right)\right]=501501\)
Hay \(\left(a+1\right)\left(a+2\right)=1003002=1001\cdot1002\)
\(\left\{\begin{matrix}a+1=1001\\a+2=1002\end{matrix}\right.\Rightarrow a=1000\)
Do đó: \(x=1000+\left(1000+1\right)=2001\)
Vậy, \(x=2001\)
