Question

Which expression correctly shows how to use the binomial theorem to determine the 4th term in the expansion of (2x²y³ + y)²?

Answer 1

Answer:

$\textsf{B)} \quad \displaystyle \sum^{7}_{k=0} \left(\frac{7!}{3!\:4!} \cdot \left(2x^2y^3\right)^{3}\left(y\right)^{4}\right)=280x^6y^{13}$

Step-by-step explanation:

$\boxed{\begin{minipage}{5cm} \underline{Binomial Theorem}\\\\ \displaystyle (a+b)^n=\sum^{n}_{k=0}\binom{n}{k} a^{n-k}b^{k} \\\\\\where \displaystyle \binom{n}{k} = \frac{n!}{(n-k)!k!}\\\end{minipage}}$

Given expression:

$(2x^2y^3+y)^7$

Therefore:

• $a=2x^2y^3$

• $b=y$

• $n=7$

Therefore:

$\displaystyle (2x^2y^3+y)^7=\sum^{7}_{k=0}\binom{7}{k} (2x^2y^3)^{7-k}(y)^{k}$

To find the 4th term in the binomial expansion, substitute k = 4 into the equation:

$\implies \displaystyle \sum^{7}_{k=0}\binom{7}{4} \left(2x^2y^3\right)^{7-4}\left(y\right)^{4}$

$\implies \displaystyle \sum^{7}_{k=0} \left(\frac{7!}{(7-4)!4!} \left(2x^2y^3\right)^{3}\left(y\right)^{4}\right)$

$\implies \displaystyle \sum^{7}_{k=0} \left(\frac{7!}{3!\:4!} \cdot \left(2x^2y^3\right)^{3}\left(y\right)^{4}\right)$

$\implies \displaystyle \sum^{7}_{k=0} \left(35 \cdot 8x^6y^9y^4\right)$

$\implies \displaystyle \sum^{7}_{k=0} \left(280x^6y^{13}\right)\right)$

Therefore, the expression that correctly shows how to use the binomial theorem to determine the 4th term in the expansion is:

$\displaystyle \sum^{7}_{k=0} \left(\frac{7!}{3!\:4!} \cdot \left(2x^2y^3\right)^{3}\left(y\right)^{4}\right)=280x^6y^{13}$