# 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.
$\sum _{x=3}^{10}3x^{3}+\sum _{x=3}^{10}4x^{2}+\sum _{x=3}^{10}5x$ 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.
$3\left[{\frac {1}{4}}10^{2}\left(10+1\right)^{2}-{\frac {1}{4}}2^{2}\left(2+1\right)^{2}\right]+4\left[{\frac {1}{6}}10\left(2\times 10+1\right)\left(10+1\right)-{\frac {1}{6}}2\left(2\times 2+1\right)\left(2+1\right)\right]$ $+5\left[{\frac {1}{2}}10(10+1)-{\frac {1}{2}}2(2+1)\right]$ 4. Now we need to perform a lot of arithmetic. This can be done by hand or utilizing a calculator.
${\frac {3}{4}}\left(100\times 121-4\times 9\right)+4{\frac {1}{6}}\left(10\times 21\times 11-10\times 3\right)$ $+{\frac {5}{2}}\left(10\times 11-2\times 3\right)$ $={\frac {3}{4}}\left(12100-36\right)+{\frac {2}{3}}\left(2310-30\right)+{\frac {5}{2}}\left(110-6\right)$ $=10828\,$ 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.