For a fixed positive integer n, if D = |n! (n+1)! (n+2)!; (n+1)! (n+2)! (n+3)!; (n+2)! (n+3)! (n+4)!

Q: Why can't I just expand the determinant directly?

Expanding a $3 \times 3$ determinant with factorials directly would lead to a massive, unmanageable expression. The beauty of this problem lies in recognizing the pattern and using row operations to simplify it before expansion.

Q: How do I know which row operations to perform?

The goal is to create an upper triangular matrix, where all elements below the main diagonal are zero. This makes the determinant simply the product of the diagonal elements. The row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$ are standard tools to achieve this.

Q: Is it okay to factor out $(n!)$ from a row if it's not explicitly written as a factor?

Yes! Remember that $(n+1)! = (n+1)n!$ and $(n+2)! = (n+2)(n+1)n!$. By factoring out $n!$ from the first row, $(n+1)!$ from the second, and $(n+2)!$ from the third, you are essentially simplifying the matrix to its core algebraic structure.

Question

For a fixed positive integer $n$, if $D = \begin{vmatrix} n! & (n + 1)! & (n + 2)! \ (n + 1)! & (n + 2)! & (n + 3)! \ (n + 2)! & (n + 3)! & (n + 4)! \end{vmatrix}$, then show that $\left[ \frac{D}{(n!)^3} - 4 ight]$ is divisible by $n$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, traveler of the JEE landscape. Today, we stand before a structure that might seem daunting at first glance: a $3 	imes 3$ determinant filled with factorials.

It is easy to feel overwhelmed by the sheer size of these numbers, but remember, in the world of mathematics, complexity is often just a mask for elegance. Our mission is to show that $\left[ \frac{D}{(n!)^3} - 4 ight]$ is divisible by $n$.

The determinant is defined as:

$$D = \begin{vmatrix} n! & (n + 1)! & (n + 2)! \ (n + 1)! & (n + 2)! & (n + 3)! \ (n + 2)! & (n + 3)! & (n + 4)! \end{vmatrix}$$

### The Pattern of Factorials

The first step is to recognize the hidden structure. We know that $(n+k)! = (n+k)(n+k-1)\dots n!$.

This is our key. By applying this, we can see that every row has a common factor. The first row has $n!$, the second has $(n+1)!$, and the third has $(n+2)!$.

### The Art of Extraction

By pulling out $n!$ from the first row, $(n+1)!$ from the second, and $(n+2)!$ from the third, we transform the matrix into something much more manageable:

$$D = (n!)(n+1)!(n+2)! \begin{vmatrix} 1 & n+1 & (n+1)(n+2) \ 1 & n+2 & (n+2)(n+3) \ 1 & n+3 & (n+3)(n+4) \end{vmatrix}$$

The first column is now entirely ones. We have stripped away the factorial complexity and revealed the underlying algebraic structure.

### The Geometry of Zeros

Now, we create zeros in the first column to simplify the determinant. We perform row operations: $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_2$.

This yields:

$$D = (n!)(n+1)!(n+2)! \begin{vmatrix} 1 & n+1 & n^2+3n+2 \ 0 & 1 & 2n+4 \ 0 & 1 & 2n+6 \end{vmatrix}$$

Performing one more operation, $R_3 	o R_3 - R_2$, we obtain:

$$D = (n!)(n+1)!(n+2)! \begin{vmatrix} 1 & n+1 & n^2+3n+2 \ 0 & 1 & 2n+4 \ 0 & 0 & 2 \end{vmatrix}$$

This is an upper triangular matrix. The determinant is simply the product of the diagonal elements: $1 	imes 1 	imes 2 = 2$.

### The Final Reveal

Now, we bring it all together:

$$D = 2 \cdot (n!) \cdot (n+1)n! \cdot (n+2)(n+1)n! = 2(n!)^3 (n+1)^2 (n+2)$$

When we substitute this into our target expression, the $(n!)^3$ terms cancel out, leaving us with $2(n+1)^2(n+2) - 4$.

Expanding this, we get:

$$2(n^3 + 4n^2 + 5n + 2) - 4 = 2n^3 + 8n^2 + 10n$$

Factoring out $n$, we get $n(2n^2 + 8n + 10)$. Since the expression inside the parentheses is an integer, the entire expression is **divisible by $n$**.

Animated Solution for Mathematics - Matrices and Determinants: For a fixed positive integer $n$ , if $D = n! (n + 1)! (n + 2)! (n + 1)! (n + 2)! (n + 3)! (n + 2)! (n + 3)! (n + 4)!$ , then show that $[\frac{D}{( n ! ) ^{3}} - 4]$ is divisible by $n$ .

Visualized Solution

Understanding the Determinant $D$

The Factorial Property

Factoring Out Common Terms

The Simplified Matrix

Applying Row Operations

Creating More Zeros

Evaluating the Final Expression for $D$

Calculating the Target Expression

Final Proof of Divisibility

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Pattern of Factorials

The Art of Extraction

The Geometry of Zeros

The Final Reveal

Similar Questions

If $D = 111 1 1 + x 1 11 1 + y$ for $x \neq = 0, y \neq = 0$ , then $D$ is

Let the three digit numbers $A 28, 3 B 9$ , and $62 C$ , where $A, B$ , and $C$ are integers between $0$ and $9$ , be divisible by a fixed integer $k$ . Show that the determinant $A 82 39 B 6 C 2$ is divisible by $k$ .

Let $a > 0, d > 0$ . Find the value of the determinant $\frac{1}{a} \frac{1}{( a + d )} \frac{1}{( a + 2 d )} \frac{1}{a ( a + d )} \frac{1}{( a + d ) ( a + 2 d )} \frac{1}{( a + 2 d ) ( a + 3 d )} \frac{1}{( a + d ) ( a + 2 d )} \frac{1}{( a + 2 d ) ( a + 3 d )} \frac{1}{( a + 3 d ) ( a + 4 d )}$ .

Let $p$ and $p + 2$ be prime numbers and let $Δ = p! (p + 1)! (p + 2)! (p + 1)! (p + 2)! (p + 3)! (p + 2)! (p + 3)! (p + 4)!$ . Then the sum of the maximum values of $α$ and $β$ , such that $p^{α}$ and $(p + 2)^{β}$ divide $Δ$ , is ________.

Show that $^{x} C_{r}^{y} C_{r}^{z} C_{r}^{x} C_{r + 1}^{y} C_{r + 1}^{z} C_{r + 1}^{x} C_{r + 2}^{y} C_{r + 2}^{z} C_{r + 2} =^{x} C_{r}^{y} C_{r}^{z} C_{r}^{x + 1} C_{r + 1}^{y + 1} C_{r + 1}^{z + 1} C_{r + 1}^{x + 2} C_{r + 2}^{y + 2} C_{r + 2}^{z + 2} C_{r + 2}$ .

Let $D_{k} = 1 n n 2 k n^{2} + n + 2 n^{2} + n 2 k - 1 n^{2} n^{2} + n + 2$ . If $\sum_{k = 1}^{n - 1} D_{k} = 96$ , then $n$ is equal to ________.

If $α$ be a repeated root of a quadratic equation $f (x) = 0$ and $A (x), B (x)$ and $C (x)$ be polynomials of degree 3, 4 and 5 respectively, then show that $A (x) A (α) A^{'} (α) B (x) B (α) B^{'} (α) C (x) C (α) C^{'} (α)$ is divisible by $f (x)$ , where prime denotes the derivatives.

Let $a, b, c$ be such that $b (a + c) \neq = 0$ . If $a - b c a + 1 b + 1 c - 1 a - 1 b - 1 c + 1 + a + 1 a - 1 (- 1)^{n + 2} a b + 1 b - 1 (- 1)^{n + 1} b c - 1 c + 1 (- 1)^{n} c = 0$ , then the value of $n$ is:

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.

For $α, β \in R$ and a natural number $n$ , let $A_{r} = r 2 r 3 r - 2 123 \frac{n ^{2}}{2} + α n^{2} - β \frac{n ( 3 n - 1 )}{2}$ . Then $2 A_{10} - A_{8}$ is equal to :