Calculate the sum of all the even numbers from 2 to 100 inclusive, 2 + 4 + 6 + ..... - Edexcel - A-Level Maths Pure - Question 10 - 2011 - Paper 1

For the series $k + 2k + 3k + ...... + 100$, the first term is $a = k$ and the last term is $l = 100$. To find the number of terms ($n$), we have: $n = \frac{100 - k}{k} + 1 = \frac{100}{k} + 1 - 1 = \frac{100}{k}$.

The sum of the series can be represented using the formula: $S_n = \frac{n}{2} (a + l)$. Substituting $n = \frac{100}{k}$, $a = k$, and $l = 100$:

$S = \frac{\frac{100}{k}}{2} (k + 100) = \frac{100}{2k} (k + 100) = \frac{100(100 + k)}{2k} = \frac{5000 + 50k}{k}.$ Thus, the sum simplifies to: $S = 50 + \frac{5000}{k}.$

The nth term of an arithmetic sequence can be calculated using the formula: $T_n = a + (n-1)d$, where $a = 2k + 1$, $d$ is the common difference, and $n = 50$. In this series, the common difference $d = (4k + 4) - (2k + 1) = 2k + 3$. Substituting values: $T_{50} = (2k + 1) + (50 - 1)(2k + 3)$ $T_{50} = 2k + 1 + 49(2k + 3) = 2k + 1 + 98k + 147 = 100k + 148.$