General term of Binomial Expansion

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 2nd term of the expansion of (2x-y)5 ? Solution: r=2, n=5,a=2x,b=-y

By using the formula above and substitution

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

=5C1(2x)4(-y)1    

= -80x4y

Sample problem 2: Find the middle term in the expansion of (x2+3y2)6. Solution: In the expansion of (a+b)n there are n+1 terms.Therefore, there are 7 terms and the middle term is the 4th term.

r=4,n=6,a= x2, b=3y2 By substitution,

4th term=6C(4-1)( x2)6-4+1(3y2)4-1

=6C(3) ( x2)3(3y2)3               

=540x6y6

Sample problem 3: Find the term involving x5 in the expansion of (2x-3y2)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 x5 is the 4th term. n=8, r=4,a=2x,b=-3y2

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

=8C3(2x)5(-3y)3

=-48384x5y3

Sample problem 4:Find the term involving x5 in the expansion of (x2-1/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. (x2)7-r+1(-1/x)r-1=x5, now we have an exponential equation, we can solve for r.

(x2)8-r(-x-1)r-1=x5 (x16-2r)(-x-r+1)=x5, 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 x5 is the 4th term, Going back to our formula above and by substitution, n=7, r=4, a=x2,y=-1/x

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

4th term= 7C(4-1)(x2)7-4+1(-1/x)4-1

4th term= 7C3(x2)4(-1/x)3

4th term= 7C3 x8(-1/x3)

4th term=-35x5

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 8th 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 x3 in the expansion of (x+2)7

5. Find the term involving x2 in the expansion of (x2-1/x)10 Answer key here

You may also like...

Leave a Reply

Your email address will not be published.

Protected by WP Anti Spam