I assume that if you are reading this post you are already familiar of binomial expansion. You can read more about binomial theorem here. This topic will focus on how to determine the rth term of binomial expansion and other problems similar to it but will not expand it. **General formula of expansion (a+b) ^{n}**

**rth term = nC(r-1)(a) ^{n-r+1}(b)^{r-1} where C is combination operator**

** Sample problem 1:** What the 2

^{nd}term of the expansion of

*? Solution: r=2, n=5,a=2x,b=-y*

**(2x-y)**^{5}By using the formula above and substitution

2^{nd} term = 5C(2-1)(2x)^{5-2+1}(-y)^{2-1}

=5C1(2x)^{4}(-y)^{1} ** **

**= -80x ^{4}y**

* Sample problem 2:* Find the middle term in the expansion of

*Solution: In the expansion of (a+b)*

**(x**^{2}+3y^{2})^{6}.^{n}there are n+1 terms.Therefore, there are 7 terms and the middle term is the 4

^{th}term.

r=4,n=6,a= x^{2}, b=3y^{2} By substitution,

4^{th} term=6C(4-1)( x^{2})^{6-4+1}(3y^{2})^{4-1}

=6C(3) ( x^{2})^{3}(3y^{2})^{3} ** **

**=540x ^{6}y^{6}**

* Sample problem 3:* Find the term involving

**x**in the expansion of

^{5}

**(2x-3y**.^{2})^{8}Solution: It is easy to recognize such term since a and b are contain different coefficients. Since the power of x is decreasing, that is the power of first term is 8, second term is 7 and so on.

We can conclude that the term involving x^{5} is the 4^{th} term. n=8, r=4,a=2x,b=-3y^{2}

4^{th} term=8C(4-1)(2x)^{8-4+1}(-3y)^{4-1}

=8C3(2x)^{5}(-3y)^{3}

**=-48384x ^{5}y^{3}**

* Sample problem 4:*Find the term involving x

^{5}in the expansion of

**(x**^{2}-1/x)^{7}Solution: Since a and b has common literal coefficient (x) we can’t do what we did in problem number 3. To solve this we will focus first to get the value of r.

From the literal coefficient part of our formula we have, **(x) ^{n-r+1}(y)^{r-1}** By substitution, we have n=7, we need to look for r. (x

^{2})

^{7-r+1}(-1/x)

^{r-1}=x

^{5}, now we have an exponential equation, we can solve for r.

(x^{2})^{8-r}(-x^{-1})^{r-1}=x^{5} (x^{16-2r})(-x^{-r+1})=x^{5}, the sign of x in the left side for now is irrelevant

By law of exponents,

16-2r+(-r+1)=5

17-3r=5

3r=12

**r=4**

therefore, term involving x^{5} is the 4^{th} term, Going back to our formula above and by substitution, n=7, r=4, a=x^{2},y=-1/x

4^{th} term= nC(r-1)(a)^{n-r+1}(b)^{r-1}

4^{th} term= 7C(4-1)(x^{2})^{7-4+1}(-1/x)^{4-1}

4^{th} term= 7C3(x^{2})^{4}(-1/x)^{3}

4^{th} term= 7C3 x^{8}(-1/x^{3})

**4 ^{th} term=-35x^{5}**

* Sample problem 5:* Find the sum of all numerical coefficients of

**(2x-3y)**^{4}.Solution: You can expand the equation and add each numerical coefficient to get the answer for this question however that is so impractical especially if you are only given 15 seconds to answer the question.

The shortest way to get the answer is to let x=1 and let y=1 and evaluate the expression.

Sum=(2x-3y)^{4},x=1 , y=1

Sum=(2(1)-3(1))^{4}

Sum=(-1)^{4}

**Sum=1**

Practice problems

Note: If the answer is too large you can keep your answer to the product of its prime factor

1. Find the 8^{th} term of (3x+2y)^{12}

2. What is the middle term in the expansion of (x-4y)^{6}?

3. Find the sum of the of all numerical coefficients of (2x+5y)^{5}

4. What is the term involving x^{3} in the expansion of (x+2)^{7}

5. Find the term involving x^{2} in the expansion of (x^{2}-1/x)^{10} Answer key here