Find $a, b, c,$ and $d$, if $$\frac{(n+3)! (n+2)!}{(n+1)! (n+1)!} = a n^3 + b n^2 + c n + d,$$ where $a, b, c,$ and $d \in \mathbb{N}$. - Leaving Cert Mathematics - Question b

Substituting $n = 1$: $\frac{(1+3)! (1+2)!}{(1+1)! (1+1)!} = \frac{4! \cdot 3!}{2! \cdot 2!} = \frac{24 \cdot 6}{2 \cdot 2} = 36.$ This gives us: $a(1)^3 + b(1)^2 + c(1) + d = a + b + c + 12 = 36.$ Simplifying gives us the equation: $a + b + c = 24.$

Substituting $n = 2$: $\frac{(2+3)! (2+2)!}{(2+1)! (2+1)!} = \frac{5! \cdot 4!}{3! \cdot 3!} = \frac{120 \cdot 24}{6 \cdot 6} = 80.$ This gives us: $a(2)^3 + b(2)^2 + c(2) + d = 8a + 4b + 2c + 12 = 80.$ Simplifying gives us the equation: $8a + 4b + 2c = 68.$ Dividing through by 4 results in: $2a + b + 0.5c = 17.$

Substituting $n = 3$: $\frac{(3+3)! (3+2)!}{(3+1)! (3+1)!} = \frac{6! \cdot 5!}{4! \cdot 4!} = \frac{720 \cdot 120}{24 \cdot 24} = 150.$ This gives us: $a(3)^3 + b(3)^2 + c(3) + d = 27a + 9b + 3c + 12 = 150.$ Simplifying gives us: $27a + 9b + 3c = 138.$ Dividing through by 3 results in: $9a + 3b + c = 46.$

Now, we have the following system of equations to solve:

1. $a + b + c = 24$.
2. $2a + b + 0.5c = 17$.
3. $9a + 3b + c = 46$.

Using substitution or elimination methods can lead us to values for $a$, $b$, $c$, and $d$. After solving, we find:

• $a = 1$,
• $b = 7$,
• $c = 16$,
• $d = 12$.