If $t_n=\frac{1}{4}(n+2)(n+3), n \in N$, then which one of the following is true?
Assertion (A) $\frac{1}{t_1}+\frac{1}{t_2}+\ldots+\frac{1}{t_{2003}}=\frac{2003}{3009}$
Reason (R) $\frac{1}{t_1}+\frac{1}{t_2}+\ldots+\frac{1}{t_n}=\frac{4 n}{(2 n+3)}$
If $A=\left[\begin{array}{ccc}1 & 2 & x \\ 4 & -1 & 7 \\ 2 & 4 & -6\end{array}\right]$ and the rank of $A$ is 2 , then the value of $x$ is equal to
$$ \left|\begin{array}{ll} 2 & 1 \\ 3 & 1 \end{array}\right|+\left|\begin{array}{cc} 1 & \frac{1}{3} \\ 3 & 1 \end{array}\right|+\left|\begin{array}{cc} \frac{1}{2} & \frac{1}{9} \\ 3 & 1 \end{array}\right|+\left|\begin{array}{cc} \frac{1}{4} & \frac{1}{27} \\ 3 & 1 \end{array}\right|+\ldots \infty= $$
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