Standard +0.3 This is a straightforward proof by induction with a geometric series. The pattern is clear (multiply by 3), the algebra is simple (factoring out 3^n), and it follows the standard induction template without requiring creative insight. Slightly above average difficulty only because it's Further Maths and requires formal proof structure.
Since true for \(n=1\), true for \(n=1,2,3\ldots\)
E1 [7]
Dependent on B1 and second E1
# Question 7:
When $n=1$, $6(3^n-1)=12$, so true for $n=1$ | B1 |
Assume true for $n=k$: $12+36+108+\ldots+(4\times3^k)=6(3^k-1)$ | E1 | Assume true for $k$
$\Rightarrow 12+36+108+\ldots+(4\times3^{k+1})=6(3^k-1)+(4\times3^{k+1})$ | M1 | Add correct next term to both sides
$=6\left[(3^k-1)+\frac{2}{3}\times3^{k+1}\right]$ | M1 | Attempt to factorise with a factor 6
$=6\left[3^k-1+2\times3^k\right]=6(3^{k+1}-1)$ | A1 | c.a.o. with correct simplification
Therefore if true for $n=k$, true for $n=k+1$ | E1 | Dependent on A1 and first E1
Since true for $n=1$, true for $n=1,2,3\ldots$ | E1 [7] | Dependent on B1 and second E1
---