Đáp án:
$ƯCLN\left[\dfrac{n(n+1)}{2}, 2n+1\right]=1$
Giải thích các bước giải:
Gọi $ƯCLN\left[\dfrac{n(n+1)}{2}, 2n+1\right]=d$
$\to \begin{cases}\dfrac{n(n+1)}{2}\ \vdots\ d\\2n+1\ \vdots\ d \end{cases}$
$\to \begin{cases}4\cdot\dfrac{n(n+1)}{2}\ \vdots\ d\\n(2n+1)\ \vdots\ d \end{cases}$
$\to \begin{cases}2n(n+1)\ \vdots\ d\\2n^2+n\ \vdots\ d \end{cases}$
$\to \begin{cases}2n^2 +2n\ \vdots\ d\\2n^2+n\ \vdots\ d \end{cases}$
$\to 2n^2 + 2n- (2n^2 + n)\ \vdots\ d$
$\to n\ \vdots\ d$
$\to 2n\ \vdots\ d$
$\to 2n +1 - 2n\ \vdots\ d$
$\to 1\ \vdots\ d$
Vậy $ƯCLN\left[\dfrac{n(n+1)}{2}, 2n+1\right]=1$
