Standard +0.3 This is a straightforward proof by induction of a summation formula with an arithmetic sequence. It requires standard induction steps (base case, inductive hypothesis, inductive step) with routine algebraic manipulation. While it's a Further Maths topic, the question follows a completely standard template with no novel insight required, making it slightly easier than average overall but typical for FP1 induction questions.
When \(n=1\), \(\frac{1}{2}n(7n-1) = 3\), so true for \(n=1\)
B1
Assume true for \(n = k\)
E1
Assume true for \(n = k\)
\(3+10+17+\cdots+(7k-4) = \frac{1}{2}k(7k-1)\)
\(\Rightarrow 3+10+17+\cdots+(7k-4)+(7(k+1)-4)\)
M1
Add \((k+1)\)th term to both sides
\(= \frac{1}{2}k(7k-1) + (7(k+1)-4)\)
\(= \frac{1}{2}[k(7k-1)+(14(k+1)-8)]\)
M1
Valid attempt to factorise
\(= \frac{1}{2}[7k^2+13k+6]\)
A1
c.a.o. with correct simplification
\(= \frac{1}{2}(k+1)(7k+6)\)
\(= \frac{1}{2}(k+1)(7(k+1)-1)\)
But this is the given result with \(k+1\) replacing \(k\). Therefore if true for \(k\) it is true for \(k+1\).
E1
Dependent on previous E1 and immediately previous A1
Since true for \(n=1\), true for \(n=1,2,3\) and so for all positive integers.
E1 [7]
Dependent on B1 and both previous E marks
# Question 6:
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $n=1$, $\frac{1}{2}n(7n-1) = 3$, so true for $n=1$ | B1 | |
| Assume true for $n = k$ | E1 | Assume true for $n = k$ |
| $3+10+17+\cdots+(7k-4) = \frac{1}{2}k(7k-1)$ | | |
| $\Rightarrow 3+10+17+\cdots+(7k-4)+(7(k+1)-4)$ | M1 | Add $(k+1)$th term to both sides |
| $= \frac{1}{2}k(7k-1) + (7(k+1)-4)$ | | |
| $= \frac{1}{2}[k(7k-1)+(14(k+1)-8)]$ | M1 | Valid attempt to factorise |
| $= \frac{1}{2}[7k^2+13k+6]$ | A1 | c.a.o. with correct simplification |
| $= \frac{1}{2}(k+1)(7k+6)$ | | |
| $= \frac{1}{2}(k+1)(7(k+1)-1)$ | | |
| But this is the given result with $k+1$ replacing $k$. Therefore if true for $k$ it is true for $k+1$. | E1 | Dependent on previous E1 and immediately previous A1 |
| Since true for $n=1$, true for $n=1,2,3$ and so for all positive integers. | E1 **[7]** | Dependent on B1 and both previous E marks |
---