Moderate -0.5 This is a standard proof by induction of a geometric series result with a straightforward base case (n=1), simple inductive step requiring basic algebra with powers of 2, and no conceptual surprises. While it's a Further Maths question, the technique is routine and the algebraic manipulation is minimal, making it easier than average for A-level but not trivial.
\(\therefore\) true for \(n = k+1\). As true for \(n=1\), and if true for \(n=k\) then true for \(n=k+1\), by induction the rule is true for all integers \(n \geq 1\)
R1
Concludes with a reasoned argument stating all three required elements
## Question 13:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Try $n=1$: LHS $= 2^{-1} = \frac{1}{2}$; RHS $= 1 - 2^{-1} = \frac{1}{2}$; LHS = RHS $\therefore$ true for $n=1$ | B1 | Demonstrates the rule is correct for $n=1$ |
| Assume true for $n=k$: $\sum_{r=1}^{k} 2^{-r} = 1 - 2^{-k}$; add $2^{-(k+1)}$ to both sides | M1 | Assumes rule true for $n=k$ **and** adds $2^{-(k+1)}$ to $1-2^{-k}$; condone incorrect/missing brackets |
| $\sum_{r=1}^{k+1} 2^{-r} = 1 - 2^{-k} + 2^{-(k+1)} = 1 + 2^{-(k+1)}(-2^1+1) = 1 - 2^{-(k+1)}$ | A1 | Correctly obtains $1 - 2^{-(k+1)}$ from $1 - 2^{-k} + 2^{-(k+1)}$ |
| $\therefore$ true for $n = k+1$. As true for $n=1$, and if true for $n=k$ then true for $n=k+1$, by induction the rule is true for all integers $n \geq 1$ | R1 | Concludes with a reasoned argument stating all three required elements |
13 Prove by induction that, for all integers $n \geq 1$
$$\sum _ { r = 1 } ^ { n } 2 ^ { - r } = 1 - 2 ^ { - n }$$
[4 marks]\\
\hfill \mbox{\textit{AQA Further AS Paper 1 2021 Q13 [4]}}