Explanation

(c)
We have, $a = 1 + 2 + 4 + 8 + ...$ upto$\ n\$terms
$= 1\left( \frac{2^{n} - 1}{2 - 1} \right) = 2^{n} - 1$
$b = 1 + 3 + 9 + ...$ upto $n$ terms=$\frac{3^{n} - 1}{2}$
and $c = 1 + 5 + 25 +$… upto $n$ terms=$\frac{5^{n} - 1}{4}$
$\therefore\left| \begin{matrix}
a & 2b & 4c \\
2 & 2 & 2 \\
2^{n} & 3^{n} & 5^{n}
\end{matrix} \right| = 2\left| \begin{matrix}
2^{n} - 1 & 3^{n} - 1 & 5^{n} - 1 \\
1 & 1 & 1 \\
2^{n} & 3^{n} & 5^{n}
\end{matrix} \right|$
$= 2\left| \begin{matrix}
2^{n} & 3^{n} & 5^{n} \\
1 & 1 & 1 \\
2^{n} & 3^{n} & 5^{n}
\end{matrix} \right|$ $\lbrack R_{1} \rightarrow R_{1} + R_{2}\rbrack$
$= 2 \times 0 = 0\ \ \lbrack\therefore$ two rows are identical]