Revision:OCR Core 2 - The binomial theorem

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Revision:OCR Core 2 - The binomial theorem

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > OCR Core 2 - The binomial theorem

1 3. The binomial theorem
2 Also See

3. The binomial theorem

Expanding $(x + y)^{n}$

If one begins with something such as:

$(x + y)^{4}$ ,

it is quick to see that this is an arduous calculation, which involves the following:

$\displaystyle (x + y)^{4} = \left[ (x + y)^{2} \right] ^{2} = \left[ x^{2} + xy + xy + y^{2} \right] ^{2} \\ = (x^{2} + 2xy + y^{2})(x^{2} + 2xy + y^{2}) \\ = x^{4} + 2x^{3}y + x^{2}y^{2} + 2x^{3}y + 4x^{2}y^{2} + 2xy^{3} + x^{2}y^{2} + 2xy^{3} + y^{4} \\ = x^{4} + 4x^{3}y + 6x^{2}y^{2} + 4xy^{3} + y^{4}$

Now, if one considers the following:

$(x + y)^{1} = 1x + 1y$

$(x + y)^{2} = 1x^{2} = 2xy + 1y^{2}$

$(x + y)^{3} = 1x^{3} + 3x^{2}y + 3xy^{2} + 1y^{3}$

$(x + y)^{4} = 1x^{4} + 4x^{3}y + 6x^{2}y^{2} + 4xy^{3} + 1y^{4}$

It is clear to see the link with Pascal's Triangle. The coefficient of each of the terms in these expansions is the same as the number on the corresponding row in Pascal's Triangle.

Also, one might note how the power of "x" falls by 1 each term, and the power of "y" increases (beginning at 0) by 1 every term.

Example

1. Expand and simplify: $(x + y + z)^{3}$ .

This is different to the examples which have been seen already, as it is a trinomial expansion, however one can do this:

$(x + y + z)^{3} = (x + (y + z))^{3}$

The row of Pascal's Triangle which will be used is the third one (1, 3, 3, 1).

Hence:

$(x + (y + z))^{3} = 1x^{3}(y + z)^{0} + 3x^{2}(y + z) + 3x(y + z)^{2} + 1x^{0}(y + z)^{3} \\ = x^{3} + 3(y + z)x^{2} + 3(y + z)^{2}x + (y + z)^{3} \\ = x^{3} + 3x^{2}y + 3x^{2}z + 3xy^{2} + 6xyz + 3z^{2} + y^{3} + 3y^{2}z + 3yz^{2} + z^{3}$ .

The binomial theorem

It has been shown (in the previous section of these notes) that the "rth" number in the "nth" row of Pascal's Triangle is:

${n \choose r}$

Hence:

$(x + y)^{n} = {n \choose 0}x^{n} + {n \choose 1}x^{n - 1}y + {n \choose 2}x^{n - 2}y^{2} + \ ... \ + {n \choose n}y^{n}$

(On calculators the value of ${n \choose r}$ is often in the "Statistics menu", or "Statistics mode", and is labelled $_{n}C_{r}$ ).

One could use the definition discussed in the previous chapter:

${n \choose r + 1} = \frac{n - r}{r + 1} {n \choose r}$ ,

however this can be a lengthy process, and therefore one should not use it during the calculation (if one has access to a calculator).

It can be used to obtain a formula for ${n \choose r}$ .

It is known that:

${n \choose 0} = 1$

Therefore:

${n \choose 0 + 1} = \frac{n - 0}{0 + 1} \times 1 = \frac{n}{1}$

${n \choose 1 + 1} = \frac{n - 1}{1 + 1} \times \frac{n}{1} = \frac{n(n - 1)}{1 \times 2}$

This pattern continues, giving:

${n \choose r} = \frac{n(n - 1)(n - 2)(n - 3) \ ... \ (n - (r - 1))}{1 \times 2 \times 3 \times \ ... \ \times r}$

One can use factorial notation:

$\displaystyle {n \choose r} = \frac{ \frac{n!}{(n - r)!} }{ r! } = \frac{n!}{r!(n - r)!}$

Often one can use the cancelling techniques discussed previously in order to make the calculations easier.

Example

1. Calculate:

${1000 \choose 998}$ .

$\displaystyle {1000 \choose 998} = \frac{1000!}{(1000 - 998)!998!} = \frac{1000!}{2! \times 998!} = \frac{999000}{2} = 499500$

2. Using the binomial theorem, expand the following:

$(x + 1)^{3}$ .

So, one follows the same technique, but one should allow "y" to equal "1":

$(x + 1)^{3} = \left[ {3 \choose 0}x^{3} \right] + \left[ 1 \times {3 \choose 1}x^2 \right] + \left[ 1^{2} \times {3 \choose 2}x \right] + \left[ 1^{3} \times {3 \choose 3} \right] \\ = x^{3} + 3x^{2} + 3x + 1$

Proving the binomial theorem

One will be aware that in Pascal's Triangle, one can obtain a number on the row below the row which one is currently considering by adding the numbers above it, for example:

${5 \choose 3} + {5 \choose 4} = {6 \choose 4}$

So, in general:

${n \choose r} + {n \choose r + 1} = {n + 1 \choose r + 1}$ .

This can be shown through the following argument:

$\displaystyle {n \choose r} + {n \choose r + 1} \\ = {n \choose r} + \frac{n - r}{r + 1} {n \choose r } \\ = \left( 1 + \frac{n - r}{r + 1} \right) {n \choose r} \\ = \left( \frac{r + 1 + (n - r)}{r + 1} \right) {n \choose r} \\ = \left( \frac{n + 1}{r + 1} \right) {n \choose r} = {n + 1 \choose r + 1}$ .