what is the second term in the binomial expansion of (2r - 3s)^12?\n-73,728r^11s\n73,728r^11s\n608,256r^10s^2…

Explanation:

Step1: Recall binomial theorem formula

The binomial expansion of $(a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}$. In the expression $(2r-3s)^{12}$, we have $a = 2r$, $b=-3s$ and $n = 12$. The $k$ - th term is $T_{k+1}=\binom{n}{k}a^{n - k}b^{k}$.

Step2: Find the second - term

For the second term, $k = 1$. First, calculate the binomial coefficient $\binom{n}{k}=\binom{12}{1}=\frac{12!}{1!(12 - 1)!}=\frac{12!}{1!11!}=12$. Then, $a^{n - k}=(2r)^{12 - 1}=(2r)^{11}=2^{11}r^{11}=2048r^{11}$, and $b^{k}=(-3s)^{1}=-3s$.

Step3: Calculate the second - term

Multiply the three parts together: $T_{2}=\binom{12}{1}(2r)^{11}(-3s)=12\times2048r^{11}\times(-3s)=-73728r^{11}s$.

Answer:

-73,728$r^{11}s$