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Matrices & Determinants: Vmatrix Vmatrix Equal

JEE Maths question with a full step-by-step solution.

Question
If Δr=∣2r−1x2n−12⋅3r−1y3n−14⋅5r−1z5n−1∣\Delta_r = \begin{vmatrix} 2^{r-1} & x & 2^n-1 \\ 2\cdot3^{r-1} & y & 3^n-1 \\ 4\cdot5^{r-1} & z & 5^n-1 \end{vmatrix}, then ∑r=1nΔr\displaystyle\sum_{r=1}^n \Delta_r is equal to:
A500500 only if n=15n = 15
B500500 only if x=y=z=0x = y = z = 0
C500500 only if x+y+z=0x + y + z = 0
DNone of thesecorrect
Solution
Step 1: Apply linearity in column 1 to sum over rr
∑r=1nΔr=∣∑r=1n2r−1x2n−1∑r=1n2⋅3r−1y3n−1∑r=1n4⋅5r−1z5n−1∣\sum_{r=1}^n \Delta_r = \begin{vmatrix} \sum_{r=1}^n 2^{r-1} & x & 2^n-1 \\ \sum_{r=1}^n 2\cdot3^{r-1} & y & 3^n-1 \\ \sum_{r=1}^n 4\cdot5^{r-1} & z & 5^n-1 \end{vmatrix}
 Step 2: Evaluate the geometric sums
∑r=1n2r−1=2n−1,∑r=1n2⋅3r−1=3n−1,∑r=1n4⋅5r−1=5n−1\sum_{r=1}^n 2^{r-1} = 2^n-1, \quad \sum_{r=1}^n 2\cdot3^{r-1} = 3^n-1, \quad \sum_{r=1}^n 4\cdot5^{r-1} = 5^n-1
 Step 3: Identify identical columns
=∣2n−1x2n−13n−1y3n−15n−1z5n−1∣=0= \begin{vmatrix} 2^n-1 & x & 2^n-1 \\ 3^n-1 & y & 3^n-1 \\ 5^n-1 & z & 5^n-1 \end{vmatrix} = 0
 Columns 1 and 3 are identical, so the sum equals 00 regardless of nn, xx, yy, zz. Answer: (4)
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