# A-level Mathematics/OCR/FP1/Summation of Series

## Summation of a Series

In Core Two we learned about arithmetic and geometric progression, but if we need to sum an arithmetic progression over a large range it can become very time consuming. There are formulae that can allow us to calculate the sum. Note that these formulae only work if we start from 1; we will see how to calculate summations from other starting points in the example below. The formulae are:

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

We also need to know this general result about summation:

$\sum_{r=1}^n ar^b = a\sum_{r=1}^n r^b$

You can see why this is true by thinking of the expanded form:

$(a \times 1^b + a \times 2^b + a \times 3^b + ... + a \times n^b) \equiv a(1^b + 2^b + 3^b + ... + n^b)$

### Example

Find the sum of the series $\sum_{x=3}^{10} 3x^3 + 4x^2 + 5x$.

1. First we need to break the summation into its three separate components.
$\sum_{x=3}^{10} 3x^3 + \sum_{x=3}^{10}4x^2 +\sum_{x=3}^{10} 5x$
2. Next we need to make them start from one. We then need to subtract the sum of the numbers not included in the summation.
$\sum_{x=1}^{10} 3x^3 - \sum_{x=1}^{2} 3x^3 + \sum_{x=1}^{10}4x^2 -\sum_{x=1}^{2}4x^2 +\sum_{x=1}^{10} 5x - \sum_{x=1}^{2} 5x$
3. Now we use the identities to calculate the individual sums. Remember to include the co-efficients.
$3\left[\frac{1}{4}10^2\left(10+1\right)^2 - \frac{1}{4}2^2\left(2+1\right)^2\right] + 4\left[\frac{1}{6}10\left(2\times10+1\right)\left(10+1\right) - \frac{1}{6}2\left(2\times2+1\right)\left(2+1\right)\right]$ $+ 5\left[ \frac{1}{2} 10(10+1) - \frac{1}{2} 2(2+1)\right]$
4. Now we need to perform a lot of arithmetic. This can be done by hand or utilizing a calculator.
$\frac{3}{4}\left(100\times121 - 4\times9\right) + 4\frac{1}{6}\left(10\times21\times11 - 10\times4\right)$ $+ \frac{5}{2}\left(10\times 11 - 2\times3\right)$
$=\frac{3}{4}\left(12100-36\right) + \frac{2}{3}\left(2310 - 30\right) + \frac{5}{2}\left(110 - 6\right)$
$=10828\,$
5. The sum of the series $\sum_{x=3}^{10} 3x^3 + 4x^2 + 5x = 10828$.

This is part of the FP1 (Further Pure Mathematics 1) module of the A-level Mathematics text.