Standard +0.3 This is a straightforward Further Maths question requiring standard summation formulas and the telescoping technique S(16 to 30) = S(1 to 30) - S(1 to 15). Part (a) is algebraic verification using given formulas, and part (b) is direct substitution. While it's Further Maths content, it's a routine textbook exercise with no novel problem-solving required, making it slightly easier than average overall.
4. (a) Use the standard results for \(\sum _ { r = 1 } ^ { n } r ^ { 3 }\) and \(\sum _ { r = 1 } ^ { n } r\) to show that
$$\sum _ { r = 1 } ^ { n } \left( r ^ { 3 } + 6 r - 3 \right) = \frac { 1 } { 4 } n ^ { 2 } \left( n ^ { 2 } + 2 n + 13 \right)$$
for all positive integers \(n\).
(b) Hence find the exact value of
$$\sum _ { r = 16 } ^ { 30 } \left( r ^ { 3 } + 6 r - 3 \right)$$
4. (a) Use the standard results for $\sum _ { r = 1 } ^ { n } r ^ { 3 }$ and $\sum _ { r = 1 } ^ { n } r$ to show that
$$\sum _ { r = 1 } ^ { n } \left( r ^ { 3 } + 6 r - 3 \right) = \frac { 1 } { 4 } n ^ { 2 } \left( n ^ { 2 } + 2 n + 13 \right)$$
for all positive integers $n$.\\
(b) Hence find the exact value of
$$\sum _ { r = 16 } ^ { 30 } \left( r ^ { 3 } + 6 r - 3 \right)$$
\hfill \mbox{\textit{Edexcel FP1 2012 Q4 [7]}}