The given image shows a question asking for the coefficients of $x^2$ and $x^3$ in the expansion of $(2x - 1)^{30}$.

We will use the Binomial Theorem to expand this expression. The Binomial Theorem states:

$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

For the given expression $(2x - 1)^{30}$:

• $a = 2x$
• $b = -1$
• $n = 30$

The general term in the expansion is:

$T_k = \binom{30}{k} (2x)^{30-k} (-1)^k$

This term simplifies to:

$T_k = \binom{30}{k} 2^{30-k} x^{30-k} (-1)^k$

Finding the coefficient of $x^2$:

We need to find the term where the power of $x$ is 2. That means $30 - k = 2$, so $k = 28$. Substituting $k = 28$ into the general term:

$T_{28} = \binom{30}{28} 2^2 x^2 (-1)^{28}$

Simplifying:

$T_{28} = \binom{30}{28} 4 x^2$

Since $\binom{30}{28} = \binom{30}{2} = \frac{30 \times 29}{2} = 435$, the coefficient of $x^2$ is:

$4 \times 435 = 1740$

Finding the coefficient of $x^3$:

We need to find the term where the power of $x$ is 3. That means $30 - k = 3$, so $k = 27$. Substituting $k = 27$ into the general term:

$T_{27} = \binom{30}{27} 2^3 x^3 (-1)^{27}$

Simplifying:

$T_{27} = \binom{30}{27} 8 x^3 (-1)$

Since $\binom{30}{27} = \binom{30}{3} = \frac{30 \times 29 \times 28}{6} = 4060$, the coefficient of $x^3$ is:

$-8 \times 4060 = -32,480$

Final Answers:

• The coefficient of $x^2$ is 1740.
• The coefficient of $x^3$ is -32,480.

Would you like further details or have any questions about this expansion?

Here are five related questions to explore:

1. How do binomial coefficients change when the exponent increases?
2. Can you find the coefficient of $x^4$ in the expansion of $(2x - 1)^{30}$?
3. What is the sum of all the coefficients in the expansion of $(2x - 1)^{30}$?
4. How would the expansion differ if the expression were $(2x + 1)^{30}$ instead?
5. What is the general form of the binomial theorem for negative exponents?

Tip: The binomial coefficients can be quickly calculated using Pascal's triangle or the formula for combinations $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.