FP2 June 2015 Q4
4.
Given that \(\displaystyle\sum_{r=1}^{n} r = \frac{1}{2}n(n + 1)\)
| Scheme | Marks |
|---|---|
| \(r^2\left(r^2 + 2r + 1\right) - \left(r^2 - 2r + 1\right)r^2\) | M1 A1 |
| \(\equiv r^4 + 2r^3 + r^2 - r^4 + 2r^3 - r^2\) or \(r^2\left(r^2 + 2r + 1 - r^2 + 2r - 1\right)\) | |
| \(\equiv 4r^3\) * | A1 |
| (3) |
Notes
M1: Multiply out brackets May remove common factor \(r^2\) first
A1: a correct statement
A1: fully correct solution which must include at least one intermediate line
ALT: Use difference of 2 squares:
M1 remove common factor and apply diff of 2 squares to rest
A1 \(r^2\left(r + 1 + r - 1\right)\left(r + 1 - (r - 1)\right)\)
\(= r^2(2r\times2)\)
A1 \(= 4r^3\)
| Scheme | Marks |
|---|---|
| \(\left(\displaystyle\sum_1^n 4r^3 =\right)\ \left(1\times2^2 - 0\right) + \left(2^2\times3^2 - 1^2\times2^2\right) + \left(3^2\times4^2 - 2^2\times3^2\right)\ldots\) \(+\left(n^2\times(n + 1)^2 - (n - 1)^2\times n^2\right)\) | M1 |
| \(= n^2(n + 1)^2\) | A1 |
| \(\displaystyle\sum_1^n r^3 = \frac{1}{4}n^2(n + 1)^2\) | A1 |
| \(\therefore \displaystyle\sum_1^n r^3 = \left(\frac{1}{2}n(n + 1)\right)^2 = \left(\sum_1^n r\right)^2\) | |
| So \(\left(1^3 + 2^3 + 3^3 + \ldots + n^3\right) = (1 + 2 + 3 \ldots + n)^2\) * | A1cso (B1 on e-PEN) |
| (4) | |
| (7 marks) |
Notes
M1: Use result to write out a list of terms; sufficient to show cancelling needed
Minimum 2 at start and 1 at end \(\displaystyle\sum_1^n 4r^3\) or \(\displaystyle\sum_1^n r^3\) need not be shown here or for next mark
A1: Correctly extracting \(n^2(n + 1)^2\) as the only remaining non-zero term.
A1: Obtaining \(\displaystyle\sum_1^n r^3 = \frac{1}{4}n^2(n + 1)^2\)
A1cso: (Shown B1 on e-PEN) for deducing the required result.
Working from either side can gain full marks
Working from both sides can gain full marks provided the working joins correctly in the middle.
If r used instead of n, penalise the final A mark.