How do you find the coefficient to the #x^6# term in the expansion of #(x-3)^9#?

1 Answer
Dec 8, 2016

Answer:

The coefficient is #color(white)1_9C_6("-"3)^(9-6)=-2268.#

Explanation:

This is a classic example of binomial expansion.

A few simple expansion of #(a+b)# to different powers:
#(a+b)^2=a^2+2ab+b^2#
#(a+b)^3=a^3+3a^2b+3ab^2+b^3#
#(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4#

You may begin to see a pattern in the coefficients. They come from Pascal's Triangle:

#"               1"#
#"            1    1"#
#"         1    2    1                "color(grey)((a+b)^2)#
#"      1   3     3   1              "color(grey)((a+b)^3)#
#"   1   4     6    4   1           "color(grey)((a+b)^4)#
#"1   5   10   10   5   1"#
#...#

Each number is the sum of the two numbers above it. The entries are also equal to #""_nC_k#, where #n# is the row number and #k# is the (diagonal) column number. (The first row is "row 0", and the first column is "column 0".)

To expand a general binomial like #(a+b)^n#, we use the following form:

#(a+b)^n=""_nC_0a^n+""_nC_1a^(n-1)b^1+""_nC_2a^(n-2)b^2+...+""_nC_nb^n#

As the terms go by, the #a# exponent counts down from #n#, while the #b# exponent counts up from #0.# So the term with #a^k# in it will be:

#color(white)1_n C_(n-k)a^kb^(n-k)#.

In this question, #n=9,# #k=6,# #a=x,# and #b="-"3.# Thus, the term with #x^6# in it is

#color(white)1_9C_(9-6)x^6("-"3)^(9-6)#
#=""_9C_3x^6("-"3)^3#
#=84x^6("-"27)#
#=-2268x^6#

So the coefficient is #-2268#.

Bonus:

We can generalize this to find the #x^k# term in a general binomial #(ax+by)^n,# where #a# and #b# are numbers. The formula will be:

#(color(white)1_nC_k*a^k*b^(n-k))x^ky^(n-k)#

and the product in the brackets will be our coefficient. For example, in the above problem, we have #a=1,b="-3"# (and #y=1#). When #n=9# and #k=6#, this gives:

#color(white)1_9C_6*(1)^6*("-3")^(9-6)#

which is the same value as before. (Note: #color(white)1_nC_k=color(white)1_nC_(n-k).#)