Moderate -0.3 This is a straightforward application of standard sum formulae (∑r² and ∑r³) with algebraic manipulation. While it requires splitting the sum, substituting formulae, and simplifying to match the given form, these are routine techniques for FP1 students with no novel problem-solving required. Slightly easier than average due to being a 'show that' question with a clear target.
5 Use standard series formulae to show that \(\sum _ { r = 1 } ^ { n } r ^ { 2 } ( 3 - 4 r ) = \frac { 1 } { 2 } n ( n + 1 ) \left( 1 - 2 n ^ { 2 } \right)\).
Appropriate use of at least one standard result; Both terms correct
\(= \frac{1}{2}n(n+1)[(2n+1)-2n(n+1)]\)
M1
Attempt to factorise using both \(n\) and \(n+1\)
\(= \frac{1}{2}n(n+1)(1-2n^2)\)
A1 [5]
Complete, convincing argument
# Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{r=1}^{n} r^2(3-4r) = 3\sum_{r=1}^{n} r^2 - 4\sum_{r=1}^{n} r^3$ | M1 | Separate into two sums involving $r^2$ and $r^3$, may be implied |
| $= \frac{3}{6}n(n+1)(2n+1) - \frac{4}{4}n^2(n+1)^2$ | M1, A1 | Appropriate use of at least one standard result; Both terms correct |
| $= \frac{1}{2}n(n+1)[(2n+1)-2n(n+1)]$ | M1 | Attempt to factorise using both $n$ and $n+1$ |
| $= \frac{1}{2}n(n+1)(1-2n^2)$ | A1 [5] | Complete, convincing argument |
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5 Use standard series formulae to show that $\sum _ { r = 1 } ^ { n } r ^ { 2 } ( 3 - 4 r ) = \frac { 1 } { 2 } n ( n + 1 ) \left( 1 - 2 n ^ { 2 } \right)$.
\hfill \mbox{\textit{OCR MEI FP1 2011 Q5 [5]}}