Standard +0.8 This is a two-part induction question from Further Maths FP1. Part (a) is a standard summation proof requiring algebraic manipulation. Part (b) involves a recurrence relation requiring substitution of both the inductive hypothesis and the recurrence formula, then simplification—more demanding than typical A-level but routine for FP1. The multi-step nature and algebraic complexity place it moderately above average difficulty.
8. (a) Prove by induction that, for \(n \in \mathbb { Z } ^ { + }\),
$$\sum _ { r = 1 } ^ { n } r ( r + 3 ) = \frac { 1 } { 3 } n ( n + 1 ) ( n + 5 )$$
(b) A sequence of positive integers is defined by
$$\begin{aligned}
u _ { 1 } & = 1 \\
u _ { n + 1 } & = u _ { n } + n ( 3 n + 1 ) , \quad n \in \mathbb { Z } ^ { + }
\end{aligned}$$
Prove by induction that
$$u _ { n } = n ^ { 2 } ( n - 1 ) + 1 , \quad n \in \mathbb { Z } ^ { + }$$
\(= \frac{1}{3}(k+1)(k+2)(k+6)\) which implies is true for \(n = k+1\)
dA1
As result is true for \(n = 1\) this implies true for all positive integers and so result is true by induction
dM1cso
(6)
(b) \(u_1 = 1^2(1-1) + 1 = 1\)
B1
(so true for \(n = 1\). Assume true for \(n = k\))
Answer
Marks
Guidance
\(u_{k+1} = k^2(k-1) + 1 + k(3k+1)\)
M1
\(= k(k^2 - k + 3k + 1) + 1 = k(k+1)^2 + 1\) which implies is true for \(n = k+1\)
A1
As result is true for \(n = 1\) this implies true for all positive integers and so result is true by induction
M1Acso
(5) [11]
Notes:
(a) First B for LHS=4 and RHS=4
First M for attempt to use \(\sum r(r+3) + u_{k+1}\)
First A for \(\frac{1}{3}(k+1)\), \(\frac{1}{3}(k+2)\) or \(\frac{1}{3}(k+6)\) as a factor before the final line
Second A dependent on first M and for any 3 of 'true for n=1' 'assume true for n=k' 'true for n=k+1', 'true for all n' (or 'true for all positive integers') seen anywhere
Third M dependent on first M and for any 3 of 'true for n=1' 'assume true for n=k' 'true for n=k+1', 'true for all n' (or 'true for all positive integers') seen anywhere
Third A for correct solution only with all statements and no errors
(b) First B for both some working and 1.
First M for \(u_{k+1} = u_k + k(3k+1)\) and attempt to substitute for \(u_k\)
First A for \(k(k+1)^2 + 1\) with some correct intermediate working and no errors seen
Second M dependent on first M and for any 3 of 'true for n=1' 'assume true for n=k' 'true for n=k+1', 'true for all n' (or 'true for all positive integers') seen anywhere
Second A for correct solution only with all statements and no errors
(a) If $n = 1$, $\sum_{r=1}^{n} r(r+3) = 1 \times 4 = 4$ and $\frac{1}{3}n(n+1)(n+5) = \frac{1}{3} \times 1 \times 2 \times 6 = 4$ | B1 |
(so true for $n = 1$. Assume true for $n = k$)
So $\sum_{r=1}^{k+1} r(r+3) = \frac{1}{3}k(k+1)(k+5) + (k+1)(k+4)$ | M1 |
$= \frac{1}{3}(k+1)[k(k+5) + 3(k+4)] = \frac{1}{3}(k+1)[k^2 + 8k + 12]$ | A1 |
$= \frac{1}{3}(k+1)(k+2)(k+6)$ which implies is true for $n = k+1$ | dA1 |
As result is true for $n = 1$ this implies true for all positive integers and so result is true by induction | dM1cso | (6)
(b) $u_1 = 1^2(1-1) + 1 = 1$ | B1 |
(so true for $n = 1$. Assume true for $n = k$)
$u_{k+1} = k^2(k-1) + 1 + k(3k+1)$ | M1 |
$= k(k^2 - k + 3k + 1) + 1 = k(k+1)^2 + 1$ which implies is true for $n = k+1$ | A1 |
As result is true for $n = 1$ this implies true for all positive integers and so result is true by induction | M1Acso | (5) [11]
**Notes:**
(a) First B for LHS=4 and RHS=4
First M for attempt to use $\sum r(r+3) + u_{k+1}$
First A for $\frac{1}{3}(k+1)$, $\frac{1}{3}(k+2)$ or $\frac{1}{3}(k+6)$ as a factor before the final line
Second A dependent on first M and for any 3 of 'true for n=1' 'assume true for n=k' 'true for n=k+1', 'true for all n' (or 'true for all positive integers') seen anywhere
Third M dependent on first M and for any 3 of 'true for n=1' 'assume true for n=k' 'true for n=k+1', 'true for all n' (or 'true for all positive integers') seen anywhere
Third A for correct solution only with all statements and no errors
(b) First B for both some working and 1.
First M for $u_{k+1} = u_k + k(3k+1)$ and attempt to substitute for $u_k$
First A for $k(k+1)^2 + 1$ with some correct intermediate working and no errors seen
Second M dependent on first M and for any 3 of 'true for n=1' 'assume true for n=k' 'true for n=k+1', 'true for all n' (or 'true for all positive integers') seen anywhere
Second A for correct solution only with all statements and no errors
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8. (a) Prove by induction that, for $n \in \mathbb { Z } ^ { + }$,
$$\sum _ { r = 1 } ^ { n } r ( r + 3 ) = \frac { 1 } { 3 } n ( n + 1 ) ( n + 5 )$$
(b) A sequence of positive integers is defined by
$$\begin{aligned}
u _ { 1 } & = 1 \\
u _ { n + 1 } & = u _ { n } + n ( 3 n + 1 ) , \quad n \in \mathbb { Z } ^ { + }
\end{aligned}$$
Prove by induction that
$$u _ { n } = n ^ { 2 } ( n - 1 ) + 1 , \quad n \in \mathbb { Z } ^ { + }$$
\hfill \mbox{\textit{Edexcel FP1 2013 Q8 [11]}}