Formulae

Summation of series can often be calculated by substituting in the following formulas:

$\sum_{r=1}^{n} 1 = n$ $\sum_{r=1}^{n} r = \frac{1}{2}n(n + 1)$ $\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n + 1)(2n + 1)$ $\sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n + 1)^2$

You can then use these equations to break down bits of a summation overr a series until it simplifies into something solvable.

Example

$$\sum_{n=1}^{20} 2n + 5$$

First, we break it down into summations for which we know the formulas:

$$\sum_{n=1}^{20} 2n + 5 = 2\sum_{n=1}^{20}n + 5\sum_{n=1}^{20}1$$

Second, we substitute the formulas in:

$$2\sum_{n=1}^{20}n + 5\sum_{r=1}^{20}1 = 2 \times \frac{1}{2} \times 20 \times 21 + 5 \times 20$$

Then, we simplify until we arrive at a valid answer:

$$2 \times \frac{1}{2} \times 20 \times 21 + 5 \times 20 = 420 + 100 = 520$$