Standard +0.3 This is a standard two-part induction question where part (a) guides students to find the formula by substituting small values of n (routine algebra), and part (b) is a straightforward induction proof with simple algebraic manipulation of factorials. The inductive step requires only basic factorial properties: (n+1)! = (n+1)·n!. This is easier than average as it's highly scaffolded and uses standard techniques without requiring insight or complex manipulation.
7. (a) It is conjectured that
$$\frac { 1 } { 2 ! } + \frac { 2 } { 3 ! } + \frac { 3 } { 4 ! } + \ldots + \frac { n - 1 } { n ! } = a - \frac { b } { n ! }$$
where \(a\) and \(b\) are constants, and \(n\) is an integer such that \(n \geq 2\). By considering particular cases, show that if the conjecture is correct then
$$a = b = 1$$
(b) Use induction to prove that, for \(n \geq 2\), the following is true
$$\frac { 1 } { 2 ! } + \frac { 2 } { 3 ! } + \frac { 3 } { 4 ! } + \ldots + \frac { n - 1 } { n ! } = 1 - \frac { 1 } { n ! }$$
7. (a) It is conjectured that
$$\frac { 1 } { 2 ! } + \frac { 2 } { 3 ! } + \frac { 3 } { 4 ! } + \ldots + \frac { n - 1 } { n ! } = a - \frac { b } { n ! }$$
where $a$ and $b$ are constants, and $n$ is an integer such that $n \geq 2$. By considering particular cases, show that if the conjecture is correct then
$$a = b = 1$$
(b) Use induction to prove that, for $n \geq 2$, the following is true
$$\frac { 1 } { 2 ! } + \frac { 2 } { 3 ! } + \frac { 3 } { 4 ! } + \ldots + \frac { n - 1 } { n ! } = 1 - \frac { 1 } { n ! }$$
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\hfill \mbox{\textit{SPS SPS FM 2024 Q7 [7]}}