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