If $Δ_r=\begin{vmatrix}2^{r-1}&2.3^{r-1}&4.5^{r-1}\\x&y&z\\2^n-1&3^n-1&5^n-1\end{vmatrix}$, then $\sum\limits_{r=1}^{n}Δ_r$ is equal to

0

Using the properties of determinants, we have

$\sum\limits_{r=1}^{n}Δ_r=\begin{vmatrix}\sum\limits_{r=1}^{n}2^{r-1}&\sum\limits_{r=1}^{n}2.3^{r-1}&\sum\limits_{r=1}^{n}4.5^{r-1}\\x&y&z\\2^n-1&3^n-1&5^n-1\end{vmatrix}$

Now, $\sum\limits_{r=1}^{n}2^{r-1}=1+2+2^2+....+2^{n-1}=1.\frac{(2^n-1)}{(2-1)}-2^n-1$

$\sum\limits_{r=1}^{n}2.3^{r-1}=2[1 +3+3^2 + ... + 3^{n-1}] = 3^n-1$

and, $\sum\limits_{r=1}^{n}4.5^{r-1}=4[1+5+5^2 +...+5^{n-1}]=5^n -1$.

$∴\sum\limits_{r=1}^{n}Δ_r=\begin{vmatrix}2^n-1&3^n-1&5^n-1\\x&y&z\\2^n-1&3^n-1&5^n-1\end{vmatrix}=0$  [∵ $R_1$ and $R_3$ are identical]