Question

If ax7yb is a term from the expansion of (x + y)12, describe how to determine its coefficient a and missing exponent b without writing the entire expansion.

Answer 1

ANSWER

a) Use the homogeneous property of the binomial expansion to find the missing exponent

b) Use the binomial theorem to find the coefficient

EXPLANATION

The given binomial expansion is:

$(x+y)^{12}$

When we compare this to

$(a + b) ^{n}$

We have

$n = 12$

Therefore the of each term in the expansion must be 12.

$\implies \: 7 + b = 12$

$b = 12 - 7$

$b = 5$

Since the coefficient of x and y are unity, we use the formula

$^{n} C_r = \frac{n!}{(n - r)!r!}$

to find the coefficient.

Where n=12 and r=5(the exponent of the y-term).

Therefore the coefficient is

$^{12} C_5= \frac{12!}{(12- 5)!5!}$

$^{12} C_5= \frac{12!}{7!5!}$

$^{12} C_5= \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7!}{7! \times 5 \times 4 \times 3 \times 2 \times 1}$

When we simplify further we get:

$^{12} C_5= 11 \times 9 \times 8 = 792$