1 Use the standard results for \(\sum _ { r = 1 } ^ { n } r\) and \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\) to show that, for all positive integers \(n\),
$$\sum _ { r = 1 } ^ { n } \left( 6 r ^ { 2 } + 2 r + 1 \right) = n \left( 2 n ^ { 2 } + 4 n + 3 \right)$$
1 Use the standard results for $\sum _ { r = 1 } ^ { n } r$ and $\sum _ { r = 1 } ^ { n } r ^ { 2 }$ to show that, for all positive integers $n$,
$$\sum _ { r = 1 } ^ { n } \left( 6 r ^ { 2 } + 2 r + 1 \right) = n \left( 2 n ^ { 2 } + 4 n + 3 \right)$$
\hfill \mbox{\textit{OCR FP1 Q1}}