Question

PLEASE HELP OR I FAIL IB MATH PLEASE in the expansion (3x-2)^12 the term in x^5 can be expressed as (12 choose r)*(3x)^p*(-2)^q

Answer 1

Answer:

$C^{12}_7(3x)^5(-2)^7$

Step-by-step explanation:

Use binomial expansion formula:

$(a+b)^n=\sum \limits _{k=0}^nC^n_ka^{n-k}b^{k}$

Then

$(3x-2)^{12}=\sum \limits_{k=0}^{12}C_k^{12}(3x)^{12-k}(-2)^k$

In the expansion $(3x-2)^{12},$ the term in $x^5$ is determined for

$12-k=5\\ \\-k=5-12\\ \\-k=-7\\ \\k=7,$

then the coefficient at $x^5$ is

$C^{12}_7(3x)^{12-7}(-2)^7=C^{12}_7(3x)^5(-2)^7$