Let Δₐ = |a-1 n 6; (a-1)² 2n² 4n-2; (a-1)³ 3n³ 3n²-3n|. Show that Σₐ₌₁ⁿ Δₐ = c, a constant.

Q: Wait, why can't I just expand the determinant directly? It looks like a standard 3x3 matrix.

You absolutely could, but that is the trap! Expanding a determinant with algebraic terms like $(a-1)^3$ and $4n-2$ will lead to a massive, messy polynomial that will take you ten minutes to simplify and likely lead to a calculation error. In JEE Advanced, if a problem looks like it requires brute force, there is almost always a property-based shortcut waiting to be discovered.

Q: How do I know for sure that I can move the summation sign inside the determinant?

Great question! The determinant is a multilinear function of its columns. If you have a determinant where only one column varies with the index $a$, you can treat the determinant as a function $f(C_1, C_2, C_3)$. Since the determinant is linear with respect to each column, $\sum f(C_1(a), C_2, C_3) = f(\sum C_1(a), C_2, C_3)$. This is a powerful tool that saves you from summing $n$ different determinants!

Q: Is the answer always zero when we see these types of problems?

Not necessarily, but it is a very common outcome in competitive exams! When you see a problem asking to prove a sum is a 'constant', and the expression involves determinants, the most elegant path is usually showing that the resulting matrix has proportional rows or columns. If the columns are proportional, the determinant is zero, and zero is indeed a constant.

Question

Let $\Delta_a = \begin{vmatrix} a - 1 & n & 6 \ (a - 1)^2 & 2n^2 & 4n - 2 \ (a - 1)^3 & 3n^3 & 3n^2 - 3n \end{vmatrix}$. Show that $\sum_{a=1}^n \Delta_a = c$, a constant.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The determinant is given by:

$$
\Delta_a = \begin{vmatrix} a - 1 & n & 6 \ (a - 1)^2 & 2n^2 & 4n - 2 \ (a - 1)^3 & 3n^3 & 3n^2 - 3n \end{vmatrix}
$$

For many students, the instinct is to panic. You see powers of $a$, you see $n$, and you see a summation sign $\sum_{a=1}^n$.

The immediate, desperate urge is to expand this $3 	imes 3$ matrix, multiply everything out, and hope for a cancellation. But stop. Take a deep breath.

In the world of JEE Advanced, brute force is rarely the intended path. It is a trap designed to consume your time and drain your energy. Today, we are going to learn how to look at this problem like a detective, not a calculator.

### The Detective's First Clue: Column Independence

Before we touch a single formula, let us observe. Look at the columns of our determinant.

Column 2 contains $n$, $2n^2$, and $3n^3$. Column 3 contains $6$, $4n-2$, and $3n^2-3n$.

Do you see the variable $a$ anywhere in these two columns? No. They are completely static. They are islands of stability in a sea of changing variables.

Only the first column, $C_1$, contains the variable $a$. This is the 'Aha!' moment.

In linear algebra, the determinant is a multilinear function of its columns. This means that if only one column varies with our summation index $a$, we do not need to sum the entire determinant $n$ times.

We can simply push the summation operator inside the determinant and apply it directly to the elements of the first column. It is like finding a secret door that bypasses the entire maze.

### The Summation Grind: Breaking Down the Elements

Now that we have established that:

$$
\sum \Delta_a = \begin{vmatrix} \sum (a-1) & n & 6 \ \sum (a-1)^2 & 2n^2 & 4n - 2 \ \sum (a-1)^3 & 3n^3 & 3n^2 - 3n \end{vmatrix}
$$

We must evaluate these sums. Let $k = a-1$. As $a$ goes from $1$ to $n$, $k$ goes from $0$ to $n-1$.

We are essentially summing the first $n-1$ natural numbers, their squares, and their cubes.

For the first element, we have:

$$
S_1 = \sum_{k=0}^{n-1} k = \frac{(n-1)n}{2}
$$

For the second element, we have:

$$
S_2 = \sum_{k=0}^{n-1} k^2 = \frac{(n-1)n(2n-1)}{6}
$$

For the third element, the sum of cubes, we have:

$$
S_3 = \sum_{k=0}^{n-1} k^3 = \left[ \frac{(n-1)n}{2} ight]^2 = \frac{n^2(n-1)^2}{4}
$$

These are standard results, but writing them out clearly is vital. Do not rush this. A single sign error here will ruin the elegance of the final result.

### The Climax: The Beauty of Proportionality

Now, look at our new determinant. It looks heavy, doesn't it? But remember, in determinant problems, whenever you see complex algebraic expressions, always look for proportionality.

Let us compare our new first column with the third column. The first element of $C_1$ is $\frac{n(n-1)}{2}$, and the first element of $C_3$ is $6$.

If we divide $C_1$ by $C_3$, we get a ratio of $\frac{n(n-1)}{12}$. Now, let us test this ratio on the second row.

We have $4n-2$. If we multiply this by our ratio $\frac{n(n-1)}{12}$, we get:

$$
\frac{2(2n-1)n(n-1)}{12} = \frac{n(n-1)(2n-1)}{6}
$$

This is exactly our $S_2$! The ratio holds. When we check the third row, the pattern continues.

We have discovered that $C_1 = \frac{n(n-1)}{12} C_3$. In the language of determinants, if one column is a scalar multiple of another, the determinant is zero.

We have proven that $\sum_{a=1}^n \Delta_a = \mathbf{0}$. Zero is a constant. We have conquered the monster.

Animated Solution for Mathematics - Matrices and Determinants: Let $Δ_{a} = a - 1 (a - 1)^{2} (a - 1)^{3} n 2 n^{2} 3 n^{3} 6 4 n - 2 3 n^{2} - 3 n$ . Show that $\sum_{a = 1}^{n} Δ_{a} = c$ , a constant.

Visualized Solution

The Determinant Summation Problem

Independence of Columns

Summation Property of Determinants

Evaluating $\sum (a - 1)$

Evaluating $\sum (a - 1)^{2}$

Evaluating $\sum (a - 1)^{3}$

Analyzing the Evaluated Determinant

Extracting Common Factors

Determinant of Proportional Columns

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Detective's First Clue

The Summation Grind

The Climax

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