Let $P=\left[\begin{array}{ccc}1& 0& 0\\ 4& 1& 0\\ 16& 4& 1\end{array}\right]$ and I be the identity matrix of order 3. If $Q=\left[q_{ij}\right]$ is a matrix such that $P^{50}-Q=I$, then $\frac{q_{31}+q_{32}}{q_{21}}$ equals

The correct option is B 103
$P=\left[\begin{array}{ccc}1& 0& 0\\ 4& 1& 0\\ 16& 4& 1\end{array}\right]$
$P^2=\left[\begin{array}{ccc}1& 0& 0\\ 8& 1& 0\\ 48& 8& 1\end{array}\right]$
$P^3=\left[\begin{array}{ccc}1& 0& 0\\ 12& 1& 0\\ 96& 12& 1\end{array}\right]$
$P^4=\left[\begin{array}{ccc}1& 0& 0\\ 16& 1& 0\\ 160& 16& 1\end{array}\right]$
For the sequence $16,48,96,160,...,$
nth term is $16\times \frac{n(n+1)}{2}$
So, $P^{50}=\left[\begin{array}{ccc}1& 0& 0\\ 200& 1& 0\\ \frac{50\times 51}{2}\times 16& 200& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 200& 1& 0\\ 20400& 200& 1\end{array}\right]$
So, $Q=P^{50}-I=\left[\begin{array}{ccc}0& 0& 0\\ 200& 0& 0\\ 20400& 200& 0\end{array}\right]$
$\frac{q_{31}+q_{32}}{q_{21}}=\frac{20400+200}{200}=103$