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Prove by mathematical induction that, for all integers `n >= 1`,
`1(1!) + 2(2!) + 3(3!) + … + n(n!) = (n + 1)! - 1`. (3 marks)
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`text(Proof)\ text{(See Worked Solutions)}`
`text(Prove true for)\ n = 1:`
`text(LHS) = 1(1!) = 1`
`text(RHS) = (1 + 1)! – 1 = 2! – 1 = 1 = text(LHS)`
`:.\ text(True for)\ \ n = 1`
`text(Assume true for)\ \ n = k:`
`1(1!) + 2(2!) + … + k(k!) = (k + 1)! – 1`
`text(Prove true for)\ \ n = k + 1`
`text(i.e.)\ \ underbrace(1(1!) + 2(2!) + … + k(k!))_((k + 1)! – 1) + (k + 1)(k + 1)! = (k + 2)! – 1`
| `text(LHS)` | `= (k + 1)! – 1 + (k + 1)(k + 1)!` |
| `= (k + 1)! [1 + (k + 1)] – 1` | |
| `= (k + 1)!(k + 2) – 1` | |
| `= (k + 2)! – 1` |
`\Rightarrow\ \ text(True for)\ \ n = k + 1`
`:.\ text(S)text(ince true for)\ \ n = 1,\ text(by PMI, true for integral)\ \ n >= 1.`
Prove by mathematical induction that for all integers `n ≥ 1`,
`1/(2!) + 2/(3!) + 3/(4!) + … + n/((n + 1)!) = 1 − 1/((n + 1)!)`. (3 marks)
--- 8 WORK AREA LINES (style=lined) ---
`text{Proof (See Worked Solutions)}`
`text(Prove for)\ n ≥ 1`
`1/(2!) + 2/(3!) + 3/(4!) + … + n/((n + 1)!) = 1 − 1/((n + 1)!)`
`text(If)\ n = 1`
| `text(LHS)` | `= 1/(2!) = 1/2` |
| `text(RHS)` | `= 1 − 1/(2!) = 1 − 1/2 = 1/2` |
`:.\ text(True for)\ n = 1`
`text(Assume true for)\ n = k`
`text(i.e.)\ \ 1/(2!) + 2/(3!) + … + k/((k + 1)!) = 1 − 1/((k + 1)!)`
`text(Prove true for)\ n = k + 1`
`text(i.e.)\ \ 1/(2!) + 2/(3!) + … + k/((k + 1)!) + (k + 1)/((k + 2)!) = 1 − 1/((k + 2)!)`
| `text(LHS)` | `= 1 − 1/((k + 1)!) + (k + 1)/((k + 2)!)` |
| `= 1 − (((k + 2) − (k + 1))/((k + 1)!(k + 2)))` | |
| `= 1 − 1/((k + 2)!)\ \ …\ text(as required)` |
`=>\ text(True for)\ n = k + 1`
`:.\ text(S)text(ince true for)\ n = 1,\ text(by PMI, true for integral)\ n ≥ 1.`