Consider the linear pattern: 5; 7; 9; .. - NSC Mathematics - Question 4 - 2021 - Paper 1

To calculate the sum of the first $n$ terms of an arithmetic sequence, we can use the formula:

$S_n = \frac{n}{2} [2a + (n-1)d]$

Here, $n = 51$, $a = 5$, and $d = 2$:

$S_{51} = \frac{51}{2} [2(5) + (51-1)(2)] = \frac{51}{2} [10 + 100] = \frac{51}{2} \cdot 110 = 51 \cdot 55 = 2805.$

The general term of the series is given by $2n + 3$. The first three terms are:

1. For $n=1$: $2(1)+3 = 5$
2. For $n=2$: $2(2)+3 = 7$
3. For $n=3$: $2(3)+3 = 9$

The last term when $n=5000$ is:

For $n=5000$: $2(5000)+3 = 10003$

Thus, the expansion is:

$\sum_{n=1}^{5000} (2n+3) = 5 + 7 + 9 + \ldots + 10003.$

We first calculate the two sums:

1. The first sum, (\sum_{n=1}^{5000} (2n+3)), can be computed as derived above:

   $S_{5000} = \sum_{n=1}^{5000} (2n + 3) = 25\,020 + 15 = 25\,020$

2. The second sum, (\sum_{n=1}^{4999} (-2n-1)):

   The general term is $-2n - 1$. For 4999 terms, the sum is:

   $\sum_{n=1}^{4999} (-2n - 1) = -2(1 + 2 + ... + 4999) - 4999$ Using the sum of the first $n$ integers:

   $= -2 \cdot \frac{4999(5000)}{2} - 4999 = -24\,999\,998 - 4999 = -25\,004\,997$

Thus the combined sum is:

$25\,020 + (-24\,999\,998) = 22\,001.$