If a = 1 + 2 + 4 + ⋯ ⋯ up to n terms, b = 1 + 3 + 9 + ⋯ ⋯ up to n terms and c = 1 +…

$a=1+2+4+8+...+$upto $n$ terms

$a=2^n-1$

$b=1+3+9+...+$ upto $n$ terms

$\Rightarrow b=\frac{3^n-1}{2}$

$\Rightarrow 2b=3^n-1$

$c=1+5+25+... t$ upto $n$ terms

$\Rightarrow c=\frac{5^n-1}{4}$

$4c=5^n-1$

$∆=\left|\begin{array}{ccc}{2}^n-1& 3^n-1& 5^n-1\\ 2& 2& 2\\ 2^n& 3^n& 5^n\end{array}\right|$

$∆=\left|\begin{array}{ccc}{2}^n& 3^n& 5^n\\ 2& 2& 2\\ 2^n& 3^n& 5^n\end{array}\right|-2\left|\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 2^n& 3^n& 5^n\end{array}\right|$ by splitting the determinant into two along the first row. Also, we know that the value of determinant is $0$ if there are two identical rows or columns.

$∆=0$