Standard +0.8 This is a standard FP2 method of differences question requiring algebraic manipulation to verify an identity, then telescoping summation to derive the sum of cubes formula. While the technique is well-practiced in Further Maths, it requires careful algebraic handling and understanding of telescoping series, placing it moderately above average difficulty.
Use result to write out list of terms; sufficient to show cancelling needed. Minimum 2 at start and 1 at end. \(\sum_1^n 4r^3\) or \(\sum_1^n r^3\) need not be shown
\(= n^2(n+1)^2\)
A1
Correctly extracting \(n^2(n+1)^2\) as the only remaining non-zero term
(Shown B1 on e-PEN) for deducing the required result. So \((1^3+2^3+3^3+\ldots+n^3)=(1+2+3\ldots+n)^2\ \ast\)
# Question 4:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $r^2(r^2+2r+1)-(r^2-2r+1)r^2$ | M1, A1 | M1: Multiply out brackets. May remove common factor $r^2$ first. A1: a correct statement |
| $\equiv r^4+2r^3+r^2-r^4+2r^3-r^2$ | | |
| $\equiv 4r^3\ \ast$ | A1 | Fully correct solution which must include at least one intermediate line |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(\sum_1^n 4r^3 =\right)\ (1\times 2^2 - 0)+(2^2\times 3^2 - 1^2\times 2^2)+(3^2\times 4^2 - 2^2\times 3^2)\ldots+(n^2\times(n+1)^2-(n-1)^2\times n^2)$ | M1 | Use result to write out list of terms; sufficient to show cancelling needed. Minimum 2 at start and 1 at end. $\sum_1^n 4r^3$ or $\sum_1^n r^3$ need not be shown |
| $= n^2(n+1)^2$ | A1 | Correctly extracting $n^2(n+1)^2$ as the only remaining non-zero term |
| $\sum_1^n r^3 = \frac{1}{4}n^2(n+1)^2$ | A1 | Obtaining $\sum_1^n r^3 = \frac{1}{4}n^2(n+1)^2$ |
| $\therefore \sum_1^n r^3 = \left(\frac{1}{2}n(n+1)\right)^2 = \left(\sum_1^n r\right)^2$ | A1 cso | (Shown B1 on e-PEN) for deducing the required result. So $(1^3+2^3+3^3+\ldots+n^3)=(1+2+3\ldots+n)^2\ \ast$ |