If $a = 1 + 2 + 4 + .......$ to n terms
$b=1+3+9+......$ to n terms
$c=1+5+25+......$ to n terms
then $\begin{vmatrix}a &2b&4c\\2&2&2\\2^n&3^n&5^n\end{vmatrix}=$
Answer & explanation
Correct answer: option 3
We have,
$a = 1 + 2 + 4 + .......$ to n terms = $2^n-1$
$b=1+3+9+......$ to n terms = $\frac{1}{2}(3^n-1)$
$c=1+5+25+......$ to n terms = $\frac{1}{4}(5^n-1)$
$∴\begin{vmatrix}a &2b&4c\\2&2&2\\2^n&3^n&5^n\end{vmatrix}$
$=\begin{vmatrix}2^n-1 &3^n-1&5^n-1\\2&2&2\\2^n&3^n&5^n\end{vmatrix}$
$=2\begin{vmatrix}2^n-1 &3^n-1&5^n-1\\1&1&1\\1&1&1\end{vmatrix}$ Applying $R_3 → R_3 - R_1$ and taking 2 common $R_2$
$=2×0=0$