9. (a) Calculate the sum of all the even numbers from 2 to 100 inclusive, 2 + 4 + 6 + ..... - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 1

In the series k + 2k + 3k + ... + 100, we recognize that:

• The first term (a) = k
• The last term (l) = 100
• The common difference (d) = k

To find the number of terms (n), use the formula for finding n:

$n = \frac{l - a}{d} + 1$

Substituting the values:

$n = \frac{100 - k}{k} + 1 = \frac{100 - k + k}{k} = \frac{100}{k}$

The sum of the series can be found using the formula for the sum of an arithmetic series:

$S_n = \frac{n}{2} (a + l)$

We have already established that:

• n = \frac{100}{k}
• a = k
• l = 100

Now substituting those into the sum formula:

$S = \frac{1}{2} \left(\frac{100}{k}\right)(k + 100)$

This simplifies to:

$S = \frac{50(100 + k)}{k}$

Separating terms gives:

$S = \frac{5000}{k} + 50$

The sequence provided can be observed for a pattern as follows:

The first term (a) is (2k + 1), and the common difference (d) can be found from:

• Second term = (4k + 4)
• First term = (2k + 1)

Thus,

$d = (4k + 4) - (2k + 1) = 2k + 3$

Using the formula for the nth term of an arithmetic sequence:

$a_n = a + (n - 1)d$

For the 50th term (n = 50):

$a_{50} = (2k + 1) + (50 - 1)(2k + 3)$

Simplifying:

$= (2k + 1) + 49(2k + 3)$ $= 2k + 1 + 98k + 147$ $= 100k + 148$

Thus, the 50th term of the sequence is (100k + 148).