Standard +0.8 This is a Further Maths FP1 question requiring algebraic manipulation of summation formulas and difference of sums. Part (a) is routine expansion and substitution of standard results. Part (b) requires the insight to express the sum from 2n+1 to 4n as a difference of two sums from 1 to 4n and 1 to 2n, then factor the resulting expression to match the given form—this involves non-trivial algebraic manipulation and is above average difficulty for FP1.
5. (a) Use the standard results for \(\sum _ { r = 1 } ^ { n } r\) and \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\) to show that
$$\sum _ { r = 1 } ^ { n } ( 2 r - 1 ) ^ { 2 } = \frac { 1 } { 3 } n \left( 4 n ^ { 2 } - 1 \right)$$
(b) Hence show that
$$\sum _ { r = 2 n + 1 } ^ { 4 n } ( 2 r - 1 ) ^ { 2 } = a n \left( b n ^ { 2 } - 1 \right)$$
where \(a\) and \(b\) are constants to be found.
M1: Attempt to use at least one standard result correctly in summing at least 2 terms of expansion of \((2r-1)^2\). A1: Correct underlined expression oe. B1: \(\sum 1 = n\)
\(= \frac{1}{3}n\big[4n^2+6n+2-6n-6+3\big]\)
M1
Attempt to factor out \(\frac{1}{3}n\) before given answer
\(= \frac{1}{3}n\big[4n^2-1\big]\)
A1
cso
Part (b)
Answer
Marks
Guidance
Answer/Working
Marks
Guidance
\(\sum_{r=2n+1}^{4n}(2r-1)^2 = f(4n) - f(2n)\) or \(f(2n+1)\)
M1
Require some use of the result in part (a) for method
5. (a) Use the standard results for $\sum _ { r = 1 } ^ { n } r$ and $\sum _ { r = 1 } ^ { n } r ^ { 2 }$ to show that
$$\sum _ { r = 1 } ^ { n } ( 2 r - 1 ) ^ { 2 } = \frac { 1 } { 3 } n \left( 4 n ^ { 2 } - 1 \right)$$
(b) Hence show that
$$\sum _ { r = 2 n + 1 } ^ { 4 n } ( 2 r - 1 ) ^ { 2 } = a n \left( b n ^ { 2 } - 1 \right)$$
where $a$ and $b$ are constants to be found.\\
\hfill \mbox{\textit{Edexcel FP1 2014 Q5 [9]}}