Question

1. what is the value of the 3rd term of the expansion $(3y - 2)^7$?$20412y^5$$35a^4b^3$$21a^5b^2$$-10206y^6$

Explanation:

Step1: Recall binomial term formula

For $(x + k)^n$, the $r$-th term is $\binom{n}{r-1}x^{n-(r-1)}k^{r-1}$

Step2: Identify values for the formula

Here, $x=3y$, $k=-2$, $n=7$, $r=3$. So $r-1=2$

Step3: Calculate binomial coefficient

$\binom{7}{2}=\frac{7!}{2!(7-2)!}=\frac{7\times6}{2\times1}=21$

Step4: Compute $(3y)$ component

$(3y)^{7-2}=(3y)^5=3^5y^5=243y^5$

Step5: Compute $(-2)$ component

$(-2)^2=4$

Step6: Multiply all components

$21\times243y^5\times4=21\times972y^5=20412y^5$

Answer:

20412$y^5$ (Option A)