Question

Write down the first three and last two terms of the binomial expansion of:

a) (3x+(2/x))^15
b) (2x-(3/x))^20

Answer 1

a)
`(3x)^(15) + 15(3x)^(14)((2)/(x)) + 105(3x)^(13)((2)/(x))^2 + ... + 15(3x)((2)/(x))^(14) + ((2)/(x))^(15)`

b)
`(2x)^(20) + 20(2x)^(19)(-(3)/(x)) + 190(2x)^(18)(-(3)/(x))^2 + ... + 20(2x)(-(3)/(x))^(19) + (-(3)/(x))^(20)`

Explanation:

The binomial expansion formula is:

`(a + b)^n = a^n + \binom{n}{1}a^(n-1) b + \binom{n}{2}a^(n-2)b^2 + ... + \binom{n}{r}a^(n-r)b^(r) + ... + b^n`
The (n r) in front of each term is the binomial coefficient. This can be calculated on a calculator using the nCr button (in this case, you'd put 15C1 for (n 1), 15C2 for (n 2), etc). This can be calculated without a calculator using this formula:

`\binom{n}{r} = \frac{{n!}}{{r!\left( {n - r} \right)!}}`

In a), a = 3x, b = 2/x and n = 15.

You can plug these into the formula above to get:

`(3x)^(15) + 15(3x)^(14)((2)/(x)) + 105(3x)^(13)((2)/(x))^2 + ... + 15(3x)((2)/(x))^(14) + ((2)/(x))^(15)`

Here's how to do this without the formula:

1. Start with the a^n
2. In the second term, you subtract 1 from a's power (in this case, 15 to 14) and add 1 to b's power (0 to 1). Then you multiply it by nCr.
3. This keeps going until you get to b^n. To check you've done it correctly, the powers of a and b in each term should add up to n. For example, in term 3 above, a's power is 13 and b's power is 2. These add to make 15.
4. Remember, a's power goes down by 1 each time, and b's power goes up by 1 each time. There is no b in the first term because b^0 = 1.

You could 'simplify' this by expanding the brackets, but this gives you massive numbers so it's probably best to leave it like this.

Applying this to b):

a = 2x, b = -3/x, n = 20

`(2x)^(20) + 20(2x)^(19)(-(3)/(x)) + 190(2x)^(18)(-(3)/(x))^2 + ... + 20(2x)(-(3)/(x))^(19) + (-(3)/(x))^(20)`