Which sum is equal to \(\sum_{k=1}^{20} (2k + 1)\)? (A) 1 + 2 + 3 + 4 + \ldots + 20 (B) 1 + 3 + 5 + 7 + \ldots + 41 (C) 3 + 4 + 5 + 6 + \ldots + 20 (D) 3 + 5 + 7 + \ldots + 41 - HSC - SSCE Mathematics Extension 1 - Question 1 - 2016 - Paper 1

To evaluate the sum (\sum_{k=1}^{20} (2k + 1)), we begin by expanding it:

$$\sum_{k=1}^{20} (2k + 1) = \sum_{k=1}^{20} 2k + \sum_{k=1}^{20} 1 = 2\sum_{k=1}^{20} k + \sum_{k=1}^{20} 1.$$

The first part, (\sum_{k=1}^{20} k), represents the sum of the first 20 natural numbers, calculated as:

$$\sum_{k=1}^{n} k = \frac{n(n + 1)}{2}.$$

For (n = 20), we have:

$$\sum_{k=1}^{20} k = \frac{20(20 + 1)}{2} = \frac{20 \times 21}{2} = 210.$$

Now substituting back:

$$2 \sum_{k=1}^{20} k = 2 \times 210 = 420.$$

The second part, (\sum_{k=1}^{20} 1), simply sums to 20 since there are 20 terms.

Adding these results together, we find:

$$420 + 20 = 440.$$

Thus, the original sum evaluates to 440, with each option examined to find that it matches option (D) as it includes all integers up to 41, which can be confirmed as a part of the resulting sum.