(a) Find the coefficient of $x^4y^2$ in the binomial expansion of $(2x+y)^6$. (b) Find the fifth term in the ascending powers of $x$ in the binomial expansion of $(2x-1)^6$.

1. Problem Description

(a) Find the coefficient of $x^4y^2$ in the binomial expansion of $(2x+y)^6$.

(b) Find the fifth term in the ascending powers of $x$ in the binomial expansion of $(2x-1)^6$.

2. Solution Steps

(a) We use the binomial theorem, which states that $(a+b)^n = \sum_{k=0}^n {n \choose k} a^{n-k}b^k$.

In our case, $a = 2x$, $b = y$, and $n = 6$.

We want the term with $x^4y^2$. This occurs when $n-k = 4$ and $k = 2$. Since $n = 6$, we have $6 - k = 4$, which means $k = 2$. Thus we look at the term with $k = 2$:

${6 \choose 2} (2x)^{6-2} (y)^2 = {6 \choose 2} (2x)^4 y^2 = {6 \choose 2} (16x^4) y^2$.

The binomial coefficient ${6 \choose 2} = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15$.

So the term is $15 (16x^4) y^2 = 240 x^4 y^2$.

The coefficient of $x^4 y^2$ is $240$.

(b) We want to find the fifth term in the ascending powers of $x$ in the expansion of $(2x-1)^6$. The general term in the expansion of $(a+b)^n$ is given by ${n \choose k} a^k b^{n-k}$. The fifth term corresponds to $k = 4$ since the first term is for $k = 0$.

In this case, $a = 2x$, $b = -1$, and $n = 6$. The fifth term is

${6 \choose 4} (2x)^4 (-1)^{6-4} = {6 \choose 4} (2x)^4 (-1)^2 = {6 \choose 4} (16x^4)(1)$.

${6 \choose 4} = \frac{6!}{4!2!} = \frac{6 \times 5}{2 \times 1} = 15$.

Thus the fifth term is $15(16x^4)(1) = 240x^4$.

3. Final Answer

(a) The coefficient of $x^4y^2$ is

2

4

0. (b) The fifth term is $240x^4$.