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**
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)
= (x

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

^{3}+ 2x^{2}y + xy^{2}+^{ }x^{2}y + 2xy^{2}+ y^{3}
= x

^{3}+ 3x^{2}y +^{ }3xy^{2}+ y^{3}
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 x

^{3}y^{0}. But y^{0}= 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)
= (x

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

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

^{4}+ 4x^{3}y +^{ }6x^{2}y^{2}+^{ }4xy^{3}+ y^{4}
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 x^{n-k}y^{k}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) x

^{3}y
b) x

^{2}y^{2}
a) See that n = 4

For x

^{3}y, the coefficient according to the theorem is
n!/(n-k)! k! = 4!/(4-1)! 1! = 4!/3!1! = 4

b) For x

^{2}y^{2}, 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 + 63x^{2}+ …, 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 + cx^{2}, find the values of a, b and c.
Q3. Find the coefficient of x

^{5}in the expansion of (x + 3) (2x – 1)^{6}.
(Hint: Expand and find the difference of coefficients of x

^{5}.)**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 + 63x

^{2}+ …
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 + cx^{2}), 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 = cx

^{2}
Q3. If you complete Q1 and Q2, I think you should be able to complete
this also.

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