THIS problem (at the bottom of the post) was sent along by a student. To help him, I provided some background knowledge/information before giving him some hints/guide as to how he could solve his problem.
FACTORIALS AND BINOMIAL THEOREM
FACTORIALS AND BINOMIAL THEOREM
The factorial on a non-negative integer n, denoted
by n!, is the product of all positive integers less than or equal to n.
For example:
4! = 4 x 3 x 2 x 1 = 24
5! = 5 x 4 x 3 x 2 x 1
3! = 3 x 2 x 1
Note that:
0! = 1
1! = 1
Note another thing also:
5! = 5 x 4!
6! = 6 x 5!
3! = 3 x 2!
And so on.
Here are some exercises in operations with
factorials.
1. 5! 3!/4! = (5 x 4!) 3!/4! = 5 x 3 x 2 = 30
(See that I rewrote 5! as (5 x 4!) to simplify
with the 4! in the denominator.)
2. 10!/(5! 8!) = (10x9)8!/(5! 8!) = 10x9/5! =
90/120 = ¾
BINOMIAL
THEOREM
When expanding terms like (x + y)3
it is not too difficult as in:
(x + y)3
= (x + y) (x
+ y) (x + y)
= (x2
+ 2xy + y2) (x + y)
= x3
+ 2x2y + xy2 + x2y + 2xy2 + y3
= x3 + 3x2y + 3xy2 + y3
Can you see that the terms are arranged with x with the highest degree
(power of three) to the one with the least power (power of zero, in the last
term)?
See that for each term following, the power of x decreases by 1 while the
power on y increases by 1.
The first term in the expanded form is really x3y0.
But y0 = 1.
See that the coefficients of the expanded expression are 1, 3, 3, 1.
Let’s go another step and expanding (x + y)4.
That is,
(x + y)4
= (x + y) 3 (x + y)
= (x3
+ 3x2y + 3xy2
+ y3) (x + y)
= (x4
+ 3x3y + 3x2y2
+ xy3) + (x3y + 3x2y2
+ 3xy3 + y4)
= x4 + 4x3y + 6x2y2 + 4xy3 + y4
See that the coefficients of the terms in the expansion are 1, 4, 6, 4,
1.
Note that coefficients of terms of binomial expansions could be taken
from the famous Pascal’s Triangle:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
Can you see how this works?
ANOTHER WAY TO GET THE COEFFICIENTS
Another way to get the coefficients is to use a theorem that makes use of
factorials.
The theorem states:
In the expansion of (x + y)n, the coefficient of the
term xn-kyk is
n!/(n-k)!
k!
Note that (n-k)! k! are in the denominator.
Let us use the binomial (x + y)4 to
see if it works.
Let’s attempt the question:
What are the coefficients of terms in the expansion
of (x + y)4:
a) x3y
b) x2y2
a) See that n = 4
For x3y, the coefficient according to the theorem is
n!/(n-k)! k! = 4!/(4-1)! 1! = 4!/3!1! = 4
b) For x2y2, the coefficient according to the
theorem is
n!/(n-k)! k! = 4!/(4-2)! 2! = 4!/2!2! = 24/4 =
6
Can you see that both answers (4 and 6) correspond
to what we know already?
NOW TO YOUR QUESTIONS:
Q1. Given that (1 + ax)n = 1 – 12x + 63x2 + …, find
a and n.
Q2. Write down in ascending powers of x, the first three term in the
expansion of (2 + ax)5. Given that the first three terms in the
expansion of (b + x) (2 + ax)5 are 96 – 176x + cx2, find
the values of a, b and c.
Q3. Find the coefficient of x5 in the expansion of (x + 3) (2x
– 1)6.
(Hint: Expand and find the difference of coefficients of x5.)
GUIDE TO YOUR ANSWERS
Q1. Use the Theorem to work out a and n. Find the coefficients of the expansion
of the binomial and compare to terms 1 – 12x + 63x2 + …
Q2. This is a bit challenging. First concentrate on (2 + ax)5.
Get the first three terms of the expansion by using the Theorem or Pascal’s Triangle
numbers).
Use those three terms to multiply by (b + 2x) as in (b + 2x)(2 + ax)5.
The first three terms that you get from this product must be COMPARED TO THE
TERMS (96 – 176x + cx2), and solve for a, b and c. Comparing means:
The first term of expansion = 96
The second term of expansion = – 176x
The third term of expansion = cx2
Q3. If you complete Q1 and Q2, I think you should be able to complete
this also.
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