\(\displaystyle S_n = \sum_{r=1}^{n}\left(4r - 1\right)\) Using the standard results for \(\displaystyle\sum_{r=1}^{n} r\) and \(\displaystyle\sum_{r=1}^{n}...

The technique here is to split a sigma expression into pieces whose sums are already known. The two standard results needed are

\[\sum_{r=1}^{n} r = \tfrac{1}{2}n\left(n + 1\right) \qquad \text{and} \qquad \sum_{r=1}^{n} 1 = n,\]

the second of these being simply the count of terms, since adding \(1\) to itself \(n\) times gives \(n\). Part (b) then uses the closed form to sum a block that does not start at \(r = 1\).

(a) Showing that \(S_{n} = n\left(2n + 1\right)\) [2 marks]

A sigma sum can be split across an addition or subtraction, and a constant multiplier can be taken outside, so

\[\sum_{r=1}^{n}\left(4r - 1\right) = 4\sum_{r=1}^{n} r - \sum_{r=1}^{n} 1 = 4 \times \tfrac{1}{2}n\left(n + 1\right) - n \qquad \textbf{M1}\]

The M1 is for the correct split and substitution of both standard results, and it survives the algebra that follows. Expanding and collecting:

\[= 2n\left(n + 1\right) - n = 2n^{2} + 2n - n = 2n^{2} + n = n\left(2n + 1\right) \qquad \textbf{A1}\]

as required. Since the result is printed, both marks are for the working, and the final factorisation is part of what the accuracy mark pays for; stopping at \(2n^{2} + n\) leaves the answer in the wrong form. Two errors dominate. The first is treating \(\displaystyle\sum_{r=1}^{n} 1\) as \(1\), which loses the \(-n\) and gives \(2n^{2} + 2n\); the summand is a constant but it is still added \(n\) times. The second is applying the formula for \(\displaystyle\sum r\) without the factor \(4\), or multiplying \(4\) by \(n\left(n + 1\right)\) rather than by half of it.

Check the closed form against a direct sum for a small value. With \(n = 3\), the terms are \(4\left(1\right) - 1 = 3\), \(4\left(2\right) - 1 = 7\) and \(4\left(3\right) - 1 = 11\), giving \(3 + 7 + 11 = 21\), while the formula gives \(3\left(7\right) = 21\). This kind of check takes seconds and catches almost every slip in a "show that" of this type.

(b) Evaluating \(\displaystyle\sum_{r=11}^{30}\left(4r - 1\right)\) [1 mark]

The formula from part (a) sums from \(r = 1\), so a sum starting at \(r = 11\) must be obtained by subtraction: take the sum of the first \(30\) terms and remove the sum of the first \(10\). The terms removed are those with \(r = 1\) up to \(r = 10\), which leaves exactly \(r = 11\) up to \(r = 30\):

\[\sum_{r=11}^{30}\left(4r - 1\right) = S_{30} - S_{10} = 30\left(2 \times 30 + 1\right) - 10\left(2 \times 10 + 1\right)\] \[= 30 \times 61 - 10 \times 21 = 1830 - 210 = 1620 \qquad \textbf{B1}\]

This is a single B mark for the correct value, though it is normally allowed as a follow-through from a wrong closed form in part (a). The error the part is built to catch is subtracting \(S_{11}\) instead of \(S_{10}\), which removes the term at \(r = 11\) as well and gives \(1830 - 11 \times 23 = 1577\). The rule is that to keep the term at \(r = k\), subtract \(S_{k-1}\).

Two checks are available. Counting: there are \(30 - 11 + 1 = 20\) terms, running from \(4\left(11\right) - 1 = 43\) to \(4\left(30\right) - 1 = 119\). Since consecutive terms differ by \(4\), this is an arithmetic progression, and its sum is the number of terms times the mean of the first and last:

\[20 \times \frac{43 + 119}{2} = 20 \times 81 = 1620,\]

which confirms the answer by an entirely independent route. A rougher magnitude check: \(20\) terms averaging about \(80\) should total roughly \(1600\), and \(1620\) fits.

Examination takeaway. Split a sigma expression term by term, take constants outside, and remember that \(\displaystyle\sum_{r=1}^{n} 1 = n\) rather than \(1\). To sum a block starting at \(r = k\), compute \(S_{n} - S_{k-1}\), and verify the count of terms as \(n - k + 1\) before trusting the result.