State the third term in the expansion of (2+x)n(2+x)^n in terms of aa and nn. Write down the first three terms in the expansion of (1−3x)5(1-3x)^5. In the expansion of (1−3x)5(2+ax)6(1-3x)^5 (2+ax)^6, the coefficient of xx is −384-384. Find the value of aa.

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Worked solution

3(ii)

1. Apply the binomial theorem to expand the first three terms

(1−3x)5=1+(51)(−3x)+(52)(−3x)2+…(1-3x)^5 = 1 + \binom{5}{1}(-3x) + \binom{5}{2}(-3x)^2 + \dots

2. Simplify the terms

=1+5(−3x)+10(9x2)=1−15x+90x2= 1 + 5(-3x) + 10(9x^2) \\ = 1 - 15x + 90x^2

Answer

1−15x+90x21 - 15x + 90x^2

3(iii)

1. Find the first two terms of the expansion of $(1-3x)^5$.

(1−3x)5=1+5(−3x)+⋯=1−15x+…(1-3x)^5 = 1 + 5(-3x) + \dots = 1 - 15x + \dots

2. Find the first two terms of the expansion of $(2+ax)^6$.

(2+ax)6=26+(61)(25)(ax)+⋯=64+6(32)ax+⋯=64+192ax+…(2+ax)^6 = 2^6 + \binom{6}{1}(2^5)(ax) + \dots = 64 + 6(32)ax + \dots = 64 + 192ax + \dots

3. Find the coefficient of $x$ in the product of the two expansions.

The terms in xx from the product (1−15x+… )(64+192ax+… )(1 - 15x + \dots)(64 + 192ax + \dots) are:
1(192ax)+(−15x)(64)=192ax−960x1(192ax) + (-15x)(64) = 192ax - 960x
So, the coefficient of xx is 192a−960192a - 960.

4. Equate the coefficient to $-384$ and solve for $a$.

192a−960=−384192a - 960 = -384
192a=576192a = 576
a=3a = 3

Answer

a=3a = 3

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