P4.3Sigma notation

• B1 for a correct general term, e.g. \(5r - 3\)
• B1 for the limits \(r = 1\) to \(29\) consistent with their general term

Worked solution: The first term is \(2\) and the common difference is \(5\), so the \(r\)th term is \(2 + 5(r - 1) = 5r - 3\).
The last term \(142\) is the 29th term, so the series is \(\sum_{r=1}^{29} (5r - 3)\).

• M1 for \(\frac{29}{2}(2 + 142)\) or \(\frac{29}{2}(2 \times 2 + 28 \times 5)\)
• A1 for \(2088\)

Worked solution: \(S_{29} = \frac{29}{2}(2 + 142) = 2088\)

• M1 for using the sum of an arithmetic series with first term \(0\) and last term \(6n - 6\) (or common difference \(6\)), or \(6 \times \frac{n(n + 1)}{2} - 6n\)
• A1* for \(3n(n - 1)\) (cso)

Worked solution: The terms form an arithmetic sequence with first term \(0\) and last term \(6n - 6\):
\(\sum_{r=1}^{n} (6r - 6) = \frac{n}{2}\left[0 + (6n - 6)\right] = \frac{n}{2}(6n - 6) = 3n(n - 1)\).

• M1 for \(\sum_{r=1}^{34} (6r - 6) - \sum_{r=1}^{11} (6r - 6)\) (or for 23 terms from \(66\) to \(198\))
• M1 for correct use of the formula with \(n = 34\) and \(n = 11\)
• A1 for \(3036\)

Worked solution: \(\sum_{r=12}^{34} (6r - 6) = \sum_{r=1}^{34} (6r - 6) - \sum_{r=1}^{11} (6r - 6)\)
\(= \frac{34}{2}(6 \times 34 - 6) - \frac{11}{2}(6 \times 11 - 6) = 3366 - 330 = 3036\)

• M1 for splitting the sum into \(\sum_{r=1}^{n} 2^{r}\), \(\sum_{r=1}^{n} 4r\) and \(\sum_{r=1}^{n} (-4)\)
• M1 for the geometric part: \(\frac{2(2^{n} - 1)}{2 - 1}\)
• M1 for the arithmetic part, e.g. \(4 \times \frac{n(n + 1)}{2}\) and \(-4n\)
• A1* for simplifying to \(2^{n+1} + 2n^{2} - 2n - 2\) (cso)

Worked solution: \(\sum_{r=1}^{n} 2^{r} = \frac{2(2^{n} - 1)}{2 - 1} = 2^{n+1} - 2\)
\(\sum_{r=1}^{n} 4r = 4 \times \frac{n(n + 1)}{2} = 2n^{2} + 2n\), and \(\sum_{r=1}^{n} (-4) = -4n\).
Adding: \(\sum_{r=1}^{n} (2^{r} + 4r - 4) = 2^{n+1} + 2n^{2} - 2n - 2\).

• M1 for evaluating their expression for two consecutive values of \(n\) either side of \(10{,}000\), e.g. \(n = 12\) and \(n = 13\)
• A1 for \(n = 13\)

Worked solution: When \(n = 12\): the sum is \(8454\) (less than \(10{,}000\)).
When \(n = 13\): the sum is \(16{,}694\) (greater than \(10{,}000\)).
The sum increases with \(n\), so the least value is \(n = 13\).