AS level

Pure Maths C1
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Binomial Expansion
This is all about expanding the difficult and time consuming expressions like (1+x)9		A Revision Aid



What you need to know is how to calculate the following
Try expanding (1+x)9 and you will soon see why a special method is required.
The formula for solving these is (1+x)n = 1 + nx + (n(n-1) / 2!)x2 + (n(n-1)(n-2) / 3!)x3...........etc
If the expression to be expanded is not 1 + ... for example (3 + 6x)7 then you need to do ..
37 (1+2x)7 and the use the binomail expansion of (1+2x)7 as normal BUT ...
....don't forget to multiply the final result by 37

Example: The coefficient of x2  in the expansion  of (1+2x)n is 40. Given that n is a positive integer, find the value of n.

(1+x)n = 1+nx+(n(n-1)/2!)x2
(1+2x)n = 1+n2x+(n(n-1)/2!)(2x)2
n2-n/2*4x2 = 40x2
n2-n = 40x2/4x2*2
n2-n = 20
n2-n-20 = 0
(n+4)(n-5) so n = -4 or 5
since n is a positive integer the answer must be 5
n = 5
The more complicated expansions in the form (a+b)n are special and a slightly different formula is required.

The Binomial Theorem formula for solving these is
(a+b)n = an + nC1an − 1b + nC2an − 2b2 + nC3an − 3b3 + ..............etc

This is complicated but will be in the formula booklet - so no need to remember it
Example : Solve (2x + 1/x)5
(2x + 1/x)5  = (2x)5 + 5C1(2x)41/x + 5C2(2x)31/x2+ 5C3(2x)21/x3 + 5C4(2x)11/x4 + 5C5(2x)01/x0 1/x5
= 32x5 + 80x3 + 80x + 40 10/x3 + 1/x5
notes
be careful if it is greater than 1 - there is a special way to deal with these.
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