Find, in ascending powers of $x$, the first three terms in the expansion of

1. Apply the binomial theorem to expand the expression up to the term in $x^2$

$(1+4x)^7 = 1 + \binom{7}{1}(4x) + \binom{7}{2}(4x)^2 + \dots$

2. Evaluate the binomial coefficients and simplify the terms

$= 1 + 7(4x) + 21(16x^2) + \dots$
$= 1 + 28x + 336x^2$

1. Apply the binomial expansion formula for the first three terms

$(2-x)^7 = 2^7 + \binom{7}{1}(2)^6(-x) + \binom{7}{2}(2)^5(-x)^2 + \dots$

2. Evaluate the binomial coefficients and powers

$= 128 + 7(64)(-x) + 21(32)(x^2) + \dots$

3. Simplify to obtain the final expression

$= 128 - 448x + 672x^2$

1. Factorise the quadratic expression in the base

$2 + 7x - 4x^2 = (1 + 4x)(2 - x)$

2. Express the given power as a product of the two expansions from parts (i) and (ii)

$(2 + 7x - 4x^2)^7 = [(1 + 4x)(2 - x)]^7 = (1 + 4x)^7(2 - x)^7$

3. Use the expansions from parts (i) and (ii) to find the terms that multiply to give $x^2$

From (i), $(1+4x)^7 = 1 + 28x + 336x^2 + \dots$
From (ii), $(2-x)^7 = 128 - 448x + 672x^2 + \dots$

$(1 + 4x)^7(2 - x)^7 \approx (1 + 28x + 336x^2)(128 - 448x + 672x^2)$

Terms in $x^2$:
$= (1)(672x^2) + (28x)(-448x) + (336x^2)(128)$

4. Calculate the coefficient of $x^2$

Coefficient of $x^2 = 672 + 28(-448) + 336(128)$
$= 672 - 12544 + 43008$
$= 31136$