1. **Problem statement:** Consider the expansion of $ (2 + x)^n $, where $ n > 3 $ and $ n \in \mathbb{Z} $. The coefficient of $ x^3 $ is four times the coefficient of $ x^5 $. Find the value of $ n $. 2. **Formula and rules:** The general term in the binomial expansion of $ (a + b)^n $ is given by: $$ T_{r+1} = \binom{n}{r} a^{n-r} b^r $$ where $ r = 0, 1, 2, ..., n $. For $ (2 + x)^n $, $ a = 2 $ and $ b = x $. 3. **Express the coefficients:** - Coefficient of $ x^3 $ is the coefficient of the term where $ r = 3 $: $$ C_3 = \binom{n}{3} 2^{n-3} $$ - Coefficient of $ x^5 $ is the coefficient of the term where $ r = 5 $: $$ C_5 = \binom{n}{5} 2^{n-5} $$ 4. **Given condition:** $$ C_3 = 4 \times C_5 $$ Substitute the expressions: $$ \binom{n}{3} 2^{n-3} = 4 \times \binom{n}{5} 2^{n-5} $$ 5. **Simplify the equation:** Divide both sides by $ 2^{n-5} $: $$ \binom{n}{3} 2^{2} = 4 \times \binom{n}{5} $$ Since $ 2^2 = 4 $, this becomes: $$ 4 \binom{n}{3} = 4 \binom{n}{5} $$ Divide both sides by 4: $$ \binom{n}{3} = \binom{n}{5} $$ 6. **Use the formula for binomial coefficients:** $$ \binom{n}{r} = \frac{n!}{r! (n-r)!} $$ So, $$ \frac{n!}{3! (n-3)!} = \frac{n!}{5! (n-5)!} $$ Cancel $ n! $: $$ \frac{1}{3! (n-3)!} = \frac{1}{5! (n-5)!} $$ Cross-multiplied: $$ 5! (n-5)! = 3! (n-3)! $$ 7. **Calculate factorials:** $$ 5! = 120, \quad 3! = 6 $$ So, $$ 120 (n-5)! = 6 (n-3)! $$ Divide both sides by 6: $$ 20 (n-5)! = (n-3)! $$ 8. **Express $ (n-3)! $ in terms of $ (n-5)! $:** $$ (n-3)! = (n-3)(n-4)(n-5)! $$ Substitute: $$ 20 (n-5)! = (n-3)(n-4)(n-5)! $$ Cancel $ (n-5)! $: $$ 20 = (n-3)(n-4) $$ 9. **Solve the quadratic equation:** $$ (n-3)(n-4) = 20 $$ $$ n^2 - 7n + 12 = 20 $$ $$ n^2 - 7n + 12 - 20 = 0 $$ $$ n^2 - 7n - 8 = 0 $$ 10. **Use quadratic formula:** $$ n = \frac{7 \pm \sqrt{(-7)^2 - 4 \times 1 \times (-8)}}{2} = \frac{7 \pm \sqrt{49 + 32}}{2} = \frac{7 \pm \sqrt{81}}{2} $$ $$ n = \frac{7 \pm 9}{2} $$ 11. **Calculate roots:** - $ n = \frac{7 + 9}{2} = \frac{16}{2} = 8 $ - $ n = \frac{7 - 9}{2} = \frac{-2}{2} = -1 $ Since $ n > 3 $, the valid solution is: $$ \boxed{8} $$