FP1 January 2012 Q6

EdexcelOld spec11 marksInduction Series

(a) Prove by induction \[\sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n + 1)^2\] (5)

(b) Using the result in part (a), show that \[\sum_{r=1}^{n} (r^3 - 2) = \frac{1}{4}n(n^3 + 2n^2 + n - 8)\] (3)

Mark scheme (a)

Scheme	Marks
\(n = 1,\ \text{LHS} = 1^3 = 1,\ \text{RHS} = \dfrac{1}{4} \times 1^2 \times 2^2 = 1\) Shows both* LHS = 1 and RHS = 1*	B1
Assume true for \(n = k\)
When \(n = k + 1\) \(\displaystyle\sum_{r=1}^{k+1} r^3 = \dfrac{1}{4}k^2(k + 1)^2 + (k + 1)^3\) Adds \((k + 1)^3\) to the given result	M1
\(= \dfrac{1}{4}(k + 1)^2[k^2 + 4(k + 1)]\) Attempt to factorise out \(\dfrac{1}{4}(k + 1)^2\) Correct expression with \(\dfrac{1}{4}(k + 1)^2\) factorised out.	dM1 A1
\(= \dfrac{1}{4}(k + 1)^2(k + 2)^2\) Must see 4 things: true for \(n = 1\), assumption true for \(n = k\), said true for \(n = k + 1\) and therefore true for all \(n\) Fully complete proof with no errors and comment. All the previous marks must have been scored.	A1cso
	(5)

See extra notes for alternative approaches

To show equivalence between \(\dfrac{1}{4}k^2(k + 1)^2 + (k + 1)^3\) and \(\dfrac{1}{4}(k + 1)^2(k + 2)^2\)

Scheme	Marks
\(\dfrac{1}{4}k^2(k + 1)^2 + (k + 1)^3 = \dfrac{1}{4}k^4 + \dfrac{3}{2}k^3 + \dfrac{13}{4}k^2 + 3k + 1\) Attempt to expand one correct expression up to a quartic	M1
\(\dfrac{1}{4}(k + 1)^2(k + 2)^2 = \dfrac{1}{4}k^4 + \dfrac{3}{2}k^3 + \dfrac{13}{4}k^2 + 3k + 1\) Attempt to expand both correct expressions up to a quartic	M1
One expansion completely correct (dependent on both M’s)	A1
Both expansions correct and conclusion	A1

To show \(\dfrac{1}{4}(k + 1)^2(k + 2)^2 - \dfrac{1}{4}k^2(k + 1)^2 = (k + 1)^3\)

Scheme	Marks
\(\dfrac{1}{4}(k + 1)^2(k + 2)^2 - \dfrac{1}{4}k^2(k + 1)^2\) Attempt to subtract	M1
\(\dfrac{1}{4}(k + 1)^2(k + 2)^2 - \dfrac{1}{4}k^2(k + 1)^2 = k^3 + 3k^2 + 3k + 1\) Obtains a cubic expression Correct expression	M1 A1
\(\dfrac{1}{4}(k + 1)^2(k + 2)^2 - \dfrac{1}{4}k^2(k + 1)^2 = (k + 1)^3\) Correct completion and comment	A1

Mark scheme (b)

Scheme	Marks
\(\sum(r^3 - 2) = \sum r^3 - \sum 2\) Attempt two sums	M1
\(= \dfrac{1}{4}n^2(n + 1)^2 - 2n\) Correct expression	A1
\(= \dfrac{n}{4}(n^3 + 2n^2 + n - 8)\) * Completion to printed answer with no errors seen.	A1
	(3)

\(\sum r^3 - \sum 2n\) is M0

Mark scheme (c)

Scheme	Marks
\(\displaystyle\sum_{r=20}^{r=50} (r^3 - 2) = \dfrac{50}{4} \times 130042 - \dfrac{19}{4} \times 7592\) \((= 1625525 - 36062)\) Attempt \(\mathrm{S}_{50} - \mathrm{S}_{20}\) or \(\mathrm{S}_{50} - \mathrm{S}_{19}\) and substitutes into a correct expression at least once. Correct numerical expression (unsimplified)	M1 A1
\(= 1\,589\,463\) cao	A1
	(3)
	(11 marks)

Scheme	Marks
\(\displaystyle\sum_{r=20}^{r=50} (r^3 - 2) = \sum_{r=20}^{r=50} r^3 - \sum_{r=20}^{r=50} (2) = \dfrac{50^2}{4} \times 51^2 - \dfrac{19^2}{4} \times 20^2 - 2 \times 31\) M1 for (\(\mathrm{S}_{50} - \mathrm{S}_{20}\) or \(\mathrm{S}_{50} - \mathrm{S}_{19}\) for cubes) – (\(2 \times 30\) or \(2 \times 31\)) A1 correct numerical expression	M1 A1
\(= 1\,589\,463\)	A1