The binomial expansion of $(2 + 5x)^4$ is given by $(2 + 5x)^4 = 4 + 160x + Bx^2 + 1000x^3 + 625x^4$ - AQA - A-Level Maths Pure - Question 5 - 2022 - Paper 2

To show this relationship, we first need to expand $(2 - 5x)^4$ using the binomial theorem:

$(2 - 5x)^4 = C(4, 0) \cdot 2^4 \cdot (-5x)^0 + C(4, 1) \cdot 2^3 \cdot (-5x)^1 + C(4, 2) \cdot 2^2 \cdot (-5x)^2 + C(4, 3) \cdot 2^1 \cdot (-5x)^3 + C(4, 4) \cdot 2^0 \cdot (-5x)^4$

Calculating each term:

1. The first term is 16, similar to the original expansion.
2. The second term is $-160x$.
3. The third term is $-1000x^3$.
4. The fourth term is $-625x^4$.

Now, when we take the difference: $(2 + 5x)^4 - (2 - 5x)^4 = (16 + 160x + 600x^2 + 1000x^3 + 625x^4) - (16 - 160x + 600x^2 - 1000x^3 + 625x^4)$

We can simplify this to find:

1. The constant terms cancel.
2. The $x$ terms: $160x + 160x = 320x$.
3. The $x^3$ terms: $1000x^3 + 1000x^3 = 2000x^3$.

Thus, we find C = 320 and D = 2000.

Using the relationship found in part (b), we can rewrite the integral as:

$\int ((2 + 5x)^4 - (2 - 5x)^4) \, dx = \int (320x + 2000x^3) \, dx$

Now applying the integration:

1. The integral of $320x$ is $\frac{320}{2} x^2 = 160x^2$.
2. The integral of $2000x^3$ is $\frac{2000}{4} x^4 = 500x^4$.

Combining these results, we obtain:

$\int ((2 + 5x)^4 - (2 - 5x)^4) \, dx = 160x^2 + 500x^4 + c$

where c is the constant of integration.