FP2 June 2011 Q4
4. Given that \[(2r + 1)^3 = Ar^3 + Br^2 + Cr + 1,\]
(a) find the values of the constants \(A\), \(B\) and \(C\). (2)
(b) Show that \[(2r + 1)^3 - (2r - 1)^3 = 24r^2 + 2\] (2)
(c) Using the result in part (b) and the method of differences, show that \[\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n + 1)(2n + 1)\] (5)
| Scheme | Marks |
|---|---|
| \((2r + 1)^3 = (2r)^3 + 3(2r)^2 + 3(2r) + 1\) | M1 |
| \(A = 8,\ B = 12,\ C = 6\) | A1 |
| (2) |
Notes
1st M1 require coefficients of 1,3,3,1 or equivalent
| Scheme | Marks |
|---|---|
| \((2r - 1)^3 = (2r)^3 - 3(2r)^2 + 3(2r) - 1\) | M1 |
| \((2r + 1)^3 - (2r - 1)^3 = 24r^2 + 2\) (*) | A1cso |
| (2) |
Notes
1st M1 require 1,-3,3,-1 or equivalent
| Scheme | Marks |
|---|---|
| \(\begin{array}{lrcl} r = 1: & 3^3 - 1^3 & = & 24 \times 1^2 + 2 \\ r = 2: & 5^3 - 3^3 & = & 24 \times 2^2 + 2 \\ & \vdots & & \vdots \\ r = n: & (2n + 1)^3 - (2n - 1)^3 & = & 24 \times n^2 + 2 \end{array}\) | M1 A1 |
| Summing: \((2n + 1)^3 - 1 = 24\sum r^2 + \left(\sum\ \right)2\) | M1 |
| \(\left(\sum 2\right) = 2n\) | B1 |
| Proceeding to \(\displaystyle\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n + 1)(2n + 1)\) | A1cso |
| (5) | |
| (9 marks) |
Notes
1st M1 for attempt with at least 1,2 and \(n\) if summing expression incorrect. RHS of display not required at this stage.
1st A1 for 1,2 and n correct.
2nd M1 require cancelling and use of \(24r^2 + 2\)
Award B1 for correct \(kn\) for their approach
2nd A1 is for correct solution only