Question

[Nguyễn Tiến Công]

1 + 3 + 5 +... + x = 1600 Tìm x

Answer 1

[Xyz OLM]

Ta có 1 + 3 + 5 + .... + x = 1600

<=> [(x - 1) : 2 + 1)] . (x + 1) : 2 = 1600

<=> \(\left(\frac{x+1}{2}\right)^2=1600\)

<=> \(\left(\frac{x+1}{2}\right)^2-40^2=0\)

<=> \(\left(\frac{x+1}{2}-40\right)\left(\frac{x+1}{2}+40\right)=0\)

<=> \(\orbr{\begin{cases}\frac{x+1}{2}-40=0\\\frac{x+1}{2}+40=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=79\left(tm\right)\\x=-81\left(\text{loại}\right)\end{cases}}\)

Vậy x = 79