$n = 1$: LHS $= 1 + p$; RHS $= 1 + p$ | B1 | Therefore result is true when $n = 1$

**Assume** inequality is true for $n = k$ (*) | |

**Multiply both sides by** $1 + p$ | E1 | and stating $1 + p \ldots 0$ before multiplying both sides by $1+p$ or justifying why inequality remains $\ldots$

$(1+p)^{k+1} \ldots (1+kp)(1+p)$ | |

**Inequality only valid since multiplication by positive number because** $1 + p \ldots 0$ | |

**Considering** $(1+kp)(1+p)$ | M1 | and attempt to multiply out

RHS $= 1 + kp + p + kp^2$ | A1 |

RHS $\ldots 1 + kp + p$ | | must have $\ldots$ correct algebra and inequalities throughout

$\Rightarrow (1+p)^{k+1} \ldots 1 + (k+1)p$ | A1 |

**Hence inequality is true when** $n = k + 1$ (**) | E1 | **6 marks** | must have (*), (**) and (***) and must have earned previous B1, M1, A1, A1 marks

**but true for** $n = 1$ **so true for** $n = 2, 3, \ldots$ **by induction** (***) | |

**(or true for all integers** $n \ldots 1$ (***)$)$ | |

**Total: 6 marks**

**Notes:**
- Statement "true for $n = 1$ may appear in conclusion such as "true for $n \ldots 1$" allowing B1 to be earned
- May write $(1+p)^{k+1} = (1+p)^k(1+p)\ldots(1+kp)(1+p)$ with justification for $\ldots$ for first E1
- May earn final E1 even if first E1 has not been earned, provided other **4 marks** are scored
- If **final** statement is "true for all $n \ldots 1$" do not award final E1

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