If [1, 1; 0, 1] · [1, 2; 0, 1] · [1, 3; 0, 1] · … … ….. [1, n-1; 0, 1] = [1, 78; 0, 1], then the inverse of [1, n; 0, 1] is:

If $$\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right] \cdot\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right] \cdot\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right] \cdot \ldots \ldots \ldots . .\left[\begin{array}{cc} 1 & n-1 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & 78 \\ 0 & 1 \end{array}\right], $$ then the inverse of $\displaystyle \left[\begin{array}{ll}1 & \mathrm{n} \\ 0 & 1\end{array}\right]$ is :