Standard +0.8 This is a multi-step Further Maths question requiring expansion of a cubic expression, manipulation of standard summation formulae, algebraic simplification to match a given form, and then application using the difference of sums technique. While the individual techniques are standard FP1 material, the algebraic manipulation is moderately demanding and the 'hence' part requires insight into splitting the range, making it above average difficulty.
2. (a) Using the formulae for \(\sum _ { r = 1 } ^ { n } r , \sum _ { r = 1 } ^ { n } r ^ { 2 }\) and \(\sum _ { r = 1 } ^ { n } r ^ { 3 }\), show that
$$\sum _ { r = 1 } ^ { n } r ( r + 1 ) ( r + 3 ) = \frac { 1 } { 12 } n ( n + 1 ) ( n + 2 ) ( 3 n + k ) ,$$
where \(k\) is a constant to be found.
(b) Hence evaluate \(\sum _ { r = 21 } ^ { 40 } r ( r + 1 ) ( r + 3 )\).
## Question 2:
### Part (a)
| Working/Answer | Marks | Notes |
|---|---|---|
| $r(r+1)(r+3) = r^3 + 4r^2 + 3r$, use $\sum r^3 + 4\sum r^2 + 3\sum r$ | M1 | Must expand and start to use at least one standard formula |
| $= \frac{1}{4}n^2(n+1)^2 + 4\left(\frac{1}{6}n(n+1)(2n+1)\right) + 3\left(\frac{1}{2}n(n+1)\right)$ | A1 A1 | First 2 A marks: one wrong term A1 A0, two wrong terms A0 A0 |
| $= \frac{1}{12}n(n+1)\{3n(n+1) + 8(2n+1) + 18\}$ or $= \frac{1}{12}n\{3n^3 + 22n^2 + 45n + 26\}$ | | |
| or $= \frac{1}{12}(n+1)\{3n^3 + 19n^2 + 26n\}$ | M1 A1 | Take out factor $kn(n+1)$ or $kn$ or $k(n+1)$ directly or from quartic |
| $= \frac{1}{12}n(n+1)\{3n^2 + 19n + 26\} = \frac{1}{12}n(n+1)(n+2)(3n+13)$ $(k=13)$ | M1 A1cao | Complete method including quadratic factor and attempt to factorise; just gives $k=13$ no working is 0 marks |
### Part (b)
| Working/Answer | Marks | Notes |
|---|---|---|
| $\sum_{21}^{40} = \sum_{1}^{40} - \sum_{1}^{20}$ | M1 | M1 for substituting 40 and 20 into answer to (a) and subtracting (NB not 40 and 21); adding terms is M0A0 |
| $= \frac{1}{12}(40\times41\times42\times133) - \frac{1}{12}(20\times21\times22\times73) = 763420 - 56210 = 707210$ | A1 cao | |
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2. (a) Using the formulae for $\sum _ { r = 1 } ^ { n } r , \sum _ { r = 1 } ^ { n } r ^ { 2 }$ and $\sum _ { r = 1 } ^ { n } r ^ { 3 }$, show that
$$\sum _ { r = 1 } ^ { n } r ( r + 1 ) ( r + 3 ) = \frac { 1 } { 12 } n ( n + 1 ) ( n + 2 ) ( 3 n + k ) ,$$
where $k$ is a constant to be found.\\
(b) Hence evaluate $\sum _ { r = 21 } ^ { 40 } r ( r + 1 ) ( r + 3 )$.\\
\hfill \mbox{\textit{Edexcel FP1 2009 Q2 [9]}}