Question

Cho A=1+5+9+...+x=501502.Tìm x.

Answer 1

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

Answer 2

nhầm x=4012012

Answer 3

giúp mk vs

Answer 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, \(x=2001\)