## Question 14:

**Case 1: n even**

Let $n = 2k$:

$(2k)^2 + 5(2k) = 4k^2 + 10k$ | M1 | Sets up proof by exploring when $n = 2k$ or $n = 2k+1$; expands to achieve a quadratic expression (may be unsimplified); condone slips e.g. $2k(2k+5) = 2k^2 + 10k$

$4k^2 + 10k = 2(2k^2 + 5k)$ which is even | A1 | Correct quadratic for either odds or evens; must show or give reason why expression is even (e.g. "2 is a factor of both terms"); expression must be fully multiplied out or factorised completely

---

**Case 2: n odd**

Let $n = 2k+1$:

$(2k+1)^2 + 5(2k+1) = 4k^2 + 14k + 6$ | dM1 | Explores both $n = 2k$ **and** $n = 2k+1$, leading to two quadratic expressions

$4k^2 + 14k + 6 = 2(2k^2 + 7k + 3)$ which is even | A1* | Requires: correct quadratic for **both** odds and evens; reason why each is even; concluding statement e.g. "Hence $n^2 + 5n$ is even for all $n(\in\mathbb{Z})$"

---

**Alternative (factorisation/logic):**

$n^2 + 5n = n(n+5)$ | SC | When $n$ is odd, $n+5$ is even so odd × even = even; when $n$ is even, $n+5$ is odd so even × odd = even; both cases score SC1010 if fully correct

---

**Alternative (proof by induction):**

Assumes true for $n = k$, substitutes $n = k+1$ into $n^2 + 5n$, expands to quadratic e.g. $k^2 + 7k + 6$ | M1 | Condone arithmetical slips

$f(k+1) = k^2 + 5k + 2k + 6 = f(k) + 2(k+3)$, which is even + even = even | A1 |

Attempts to substitute $n=1 \Rightarrow 1^2 + 5\times1 = 6$ (true) | dM1 | Condone arithmetical slips

States: true for $n=1$; if true for $n=k$ then true for $n=k+1$; therefore true for all $n(\in\mathbb{Z})$ | A1* |

**(4 marks total)**