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

From Wikibooks, open books for an open world
Jump to: navigation, search

Summation of a Series[edit]

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)


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)
  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.