## Binomial Expansion When Looking For Just One Term.

Hello again all.

Finding a particular term of an expanded binomial can be tricky. We could just expand the entire binomial but that will be far too time consuming. Instead we will use the formula seen below. The variables n, X, and Y will always be given by the binomial. the "k" is the tricky one. If you can find "k" you will be able to find the particular term that you are looking for.

**t _{k+1} = _{n}C_{k} * (X)^{n-k} * (Y)^{k} ,**

where n is the exponent on the binomial, X is the first term of the binomial, Y is the second term of the binomial, and "k" is the SOB that you will always be trying to find.

I will show you 3 different particular terms that you could be asked to find and how to find "k" in each case.

**The nth term**

(2x-3)^{5}

let's find the 4th term

we have....

k=?

n=5

X=2x

Y=-3

we would set the subscript on t in the formual equal to the term # that we want:

k + 1 = 4

-1 -1

k = 3

Now we can substitute into the formula and find the 4th term........

t_{3+1} = t_{4} = _{5}C_{3} * (2x)^{5-3} * (-3)^{3}

t_{4} = 10 * 4x^{2} *-27

t_{4} = -1080x^{2}

**The middle term**

(x-2y)^{6}

note: in order to have a middle term there must be an odd # of terms, and therefore the exponent on the binomial must be even.

k = ?, (always a problem)

n = 6

X = x

Y = -2y

let's find "k"......

remember from expanding the entire binomial in the last blog that the k's on the right side of the C's started at 0 and increased to the binomial exponent.

the k's in this expansion wil be: 0 1 2 3 4 5 6

the middle "k" is 3

(note: this result shows that if you are looking for the middle term the "k" will always be half of n)

let's find the middle term....

t_{3+1} = _{6}C_{3} * (x)^{6-3} * (-2y)^{3}

t_{4} = 20 * x^{3} * -8y^{3}

t_{4} = -160x^{3}y^{3}

**The constant term**

(3x^{3} + 2/x^{3})^{6}

note: the constant term is usually the last one of an expansion, but if there are variables in both terms of the binomial the constant will be in a different postion if it exists at all

k = ? , as always

n = 6

X = 3x^{3}

Y = 2/x^{3}

let's find "k".......

it is important to realize that in order for a term to be constant it cannot have any x's in it. In every term of the expansion there will be x^{?} in the top and x^{?} in the bottom.

ie x^{4}/x^{2}

Thinking back to your exponent rules, you would subtract the 2 exponents and place the result on a single base, in this case x.

Therefore, the only way to get rid of the x's in one of the terms is to have the exponents subtract to 0, in another words the exponents on the top and bottom must be the same, since x^{0} is 1.

With that in mind we will set the exponents from the 2 terms equal to one another and solve for "k".

3(n-k) = 3(k)

the 3's are exponents from the binomial, and the (n-k) and k are from the formula directly.

3n - 3k = 3k

3(6) -3k = 3k

+3k +3k

18 = 6k

k=3

now let's find the constant term .........

t_{3+1} = t_{4} = _{6}C_{3} * (3x^{3})^{6-3} * (2/x^{3})^{3}

t_{4} = 20 * 27x^{9} *8/x^{9} the x^{9} cancels top and bottom

t_{4} = 4320

The 3 examples covered in this BLOG are the 3 most common types asked on tests but there are others that can be asked as well.

Just remember it is always "k" that you are after. Once you have him you simply plug n, k, X, Y into the formula and do the math.

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