| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove summation formula |
| Difficulty | Standard +0.8 Part (a) is a standard induction proof with straightforward algebra, typical of Core Pure AS. Part (b) requires manipulating the summation formula using the result from (a) and factorizing a quartic expression into the given form, which demands more algebraic insight and problem-solving beyond routine induction practice. |
| Spec | 4.01a Mathematical induction: construct proofs4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(n=1\): lhs \(= 1(2)(3) = 6\), rhs \(= \frac{1}{2}(1)(2)^2(3) = 6\), true for \(n=1\) | B1 | Substitutes \(n=1\) into both sides to show both equal 6 |
| Assume true for \(n=k\): \(\sum_{r=1}^{k} r(r+1)(2r+1) = \frac{1}{2}k(k+1)^2(k+2)\) | M1 | Makes statement assuming result true for some \(n = k\) |
| \(\sum_{r=1}^{k+1} r(r+1)(2r+1) = \frac{1}{2}k(k+1)^2(k+2) + (k+1)(k+2)(2k+3)\) | M1 | Adds the \((k+1)\)th term to assumed result |
| \(= \frac{1}{2}(k+1)(k+2)\left[k(k+1) + 2(2k+3)\right]\) | dM1 | Factorises out \(\frac{1}{2}(k+1)(k+2)\) |
| \(= \frac{1}{2}(k+1)(k+2)\left[k^2+5k+6\right] = \frac{1}{2}(k+1)(k+2)^2(k+3)\) | A1 | Reaches correct required expression for \(n = k+1\) |
| If true for \(n=k\) then true for \(n=k+1\), and true for \(n=1\), therefore true for all positive integers \(n\) | A1 | Correct conclusion conveying all key points |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sum_{r=n}^{2n} r(r+1)(2r+1) = \frac{1}{2}(2n)(2n+1)^2(2n+2) - \frac{1}{2}(n-1)n^2(n+1)\) | M1 | Realises \(\sum_{r=1}^{2n} - \sum_{r=1}^{n-1}\) is required and uses result from (a) |
| \(= \frac{1}{2}n(n+1)\left[4(2n+1)^2 - n(n-1)\right]\) | M1 | Attempts to factorise by \(\frac{1}{2}n(n+1)\) |
| \(= \frac{1}{2}n(n+1)(15n^2+17n+4) = \frac{1}{2}n(n+1)(3n+1)(5n+4)\) | A1 | Correct expression or correct values |
# Question 8:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $n=1$: lhs $= 1(2)(3) = 6$, rhs $= \frac{1}{2}(1)(2)^2(3) = 6$, true for $n=1$ | B1 | Substitutes $n=1$ into both sides to show both equal 6 |
| Assume true for $n=k$: $\sum_{r=1}^{k} r(r+1)(2r+1) = \frac{1}{2}k(k+1)^2(k+2)$ | M1 | Makes statement assuming result true for some $n = k$ |
| $\sum_{r=1}^{k+1} r(r+1)(2r+1) = \frac{1}{2}k(k+1)^2(k+2) + (k+1)(k+2)(2k+3)$ | M1 | Adds the $(k+1)$th term to assumed result |
| $= \frac{1}{2}(k+1)(k+2)\left[k(k+1) + 2(2k+3)\right]$ | dM1 | Factorises out $\frac{1}{2}(k+1)(k+2)$ |
| $= \frac{1}{2}(k+1)(k+2)\left[k^2+5k+6\right] = \frac{1}{2}(k+1)(k+2)^2(k+3)$ | A1 | Reaches correct required expression for $n = k+1$ |
| If true for $n=k$ then true for $n=k+1$, and true for $n=1$, therefore true for all positive integers $n$ | A1 | Correct conclusion conveying all key points |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sum_{r=n}^{2n} r(r+1)(2r+1) = \frac{1}{2}(2n)(2n+1)^2(2n+2) - \frac{1}{2}(n-1)n^2(n+1)$ | M1 | Realises $\sum_{r=1}^{2n} - \sum_{r=1}^{n-1}$ is required and uses result from (a) |
| $= \frac{1}{2}n(n+1)\left[4(2n+1)^2 - n(n-1)\right]$ | M1 | Attempts to factorise by $\frac{1}{2}n(n+1)$ |
| $= \frac{1}{2}n(n+1)(15n^2+17n+4) = \frac{1}{2}n(n+1)(3n+1)(5n+4)$ | A1 | Correct expression or correct values |
---\begin{enumerate}
\item (a) Prove by induction that, for all positive integers $n$,
\end{enumerate}
$$\sum _ { r = 1 } ^ { n } r ( r + 1 ) ( 2 r + 1 ) = \frac { 1 } { 2 } n ( n + 1 ) ^ { 2 } ( n + 2 )$$
(b) Hence, show that, for all positive integers $n$,
$$\sum _ { r = n } ^ { 2 n } r ( r + 1 ) ( 2 r + 1 ) = \frac { 1 } { 2 } n ( n + 1 ) ( a n + b ) ( c n + d )$$
where $a$, $b$, $c$ and $d$ are integers to be determined.
\hfill \mbox{\textit{Edexcel CP AS 2021 Q8 [9]}}