Math AA SL

Using Pascal’s triangle or the Binomial Theorem formula:

(x + 2)^{3} = x^{3} + (3)(x^{2})(2) + (3)(x)(2^{2}) + 2^{3} =

\boldsymbol{{\color{Purple} = x^{3} + 6x^{2} + 12x + 8}}

.

(1 - 2p)^{4} = 1^{4} + 4(1)^{3}(-2p)^{1} + 6(1)^{2}(-2p)^{2} + 4(1)^{1}(-2p)^{3} + (-2p)^{4}

\boldsymbol{{\color{Purple} 16p^{4} - 32p^{3} + 24p^{2} - 8p + 1}}

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(a)

The number of terms is one more than the expoinent in the expansion, so:

\boldsymbol{{\color{Purple} 7\: \textbf{terms}}}

(b)

Using the general term of the Binomial Theorem formula, for the x^{4} term we get:

^{6}C_{2} \times x^{6 - 2} \times 2^{2}

Using the formula for ^{n}C_{r}, we get that:

^{6}C_{2} = \frac{6!}{2!\times (6-2)!} = \frac{6 \times 5 \times 4!}{2!\times 4!} = \frac{6 \times 5}{2} = 15

So for the x^{4} term we get:

15 \times x^4 \times 4 = 60x^{4}

Therefore the coefficient of the x^{4} term is:

\boldsymbol{{\color{Purple} 60}}

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Since we are multiplying the expansion of (2x - 3)^{8} by x, to find the the coefficient of the x^{5} term in the entire expansion, we need to find the coefficient of the x^{4} term in the expansion of (2x - 3)^{8}.

Using the general term of the Binomial Theorem formula we get:

^{8}C_{4} \times (2x)^{8 - 4} \times (-3)^{4}

Using the calculator to work further we get:

70 \times 16x^{4} \times 81 = 90720x^{4}

Therefore the coefficient of the x^{5} term is:

\boldsymbol{{\color{Purple} 90720}}

.

Using the binomial expansion formula:

^{8}\textup{C}_{2} \times (2x)^{2} \times a^{6} = 224x^{2}

Calculating ^{8}\textup{C}_{2}:

\frac{8!}{2!(8-2)!} = \frac{8 \times 7 \times 6!}{2! \times 6!} = \frac{56}{2} = 28

So we get:

28 \times 4x^{2} \times a^{6} = 224x^{2}

a^{6} = 2

\boldsymbol{{\color{Purple} a = \pm \sqrt[6]{2}}}

.

(a)

The number of terms is one more than the exponent in the expansion, so:

\boldsymbol{{\color{Purple} 13\: \textbf{terms}}}

(b)

The constant term in the expansion is the term that does not contain x, hence the term where the exponent of the 2x term is the same as the exponent of the - \frac{3}{x} term.

Using the general term of the Binomial Theorem formula, this term can be expressed as follows:

^{12}C_{6} \times (2x)^{12 - 6} \times \left ( -\frac{3}{x} \right )^{6}

Using the calculator to work further we get:

924 \times 64x^{6} \times \frac{729}{x^{6}}

Simplifying, we get that the constant term is:

\boldsymbol{{\color{Purple} 43110144}}

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Using the Binomial Theorem formula, we can write that the general term in this expansion is:

^{8}C_{r} \times \left ( \frac{b}{2x} \right )^{8 - r} \times \left (  - 3x^{3} \right )^{r} =\,  ^{8}C_{r} \times\left ( \frac{b}{2} \right )^{8 - r} \times \left ( \frac{1}{x} \right )^{8 - r} \times \left (  - 3 \right )^{r} \times \left (  x^{3} \right )^{r}

Let’s focus on the power of x to find the value of r for the constant term.

The constant term in the expansion is the term that does not contain x, in other words, it is the term where the power of x is zero.

So we can write that for the constant term:

\left ( \frac{1}{x} \right )^{8 - r} \times \left (  x^{3} \right )^{r} = x^{0}

Working further we get:

\left ( x^{-1} \right )^{8 - r} \times \left (  x^{3} \right )^{r} = x^{0}

x^{r - 8} \times x^{3r} = x^{0}

x^{4r - 8} = x^{0}

4r - 8 = 0 \Rightarrow r = 2

Using this result and the information that the constant term is 1032192, we can write that:

^{8}C_{2} \times \left ( \frac{b}{2x} \right )^{8 - 2} \times \left (  - 3x^{3} \right )^{2} = 1032192

Using the calculator, working further and simplifying, we get that:

28 \times \frac{b^{6}}{64x^{6}} \times 9x^{6} = 1032192

b^{6} = 262144

\boldsymbol{{\color{Purple} b = \pm 8}}

.

(a)

Using the general term of the Binomial Theorem formula, for the x^{3} term we get:

^{7}C_{4} \times (3x)^{7 - 4} \times (-5)^{4}

Using the calculator to work further we get:

35 \times 27x^{3} \times 625

Therefore the coefficient of the x^{3} term is:

\boldsymbol{{\color{Purple} 590625}}

(b)

We can rewrite this expansion in the following way:

(2x - 3)(3x - 5)^{7} = 2x(3x - 5)^{7} - 3(3x - 5)^{7}

To get the x^{4} term from 2x(3x - 5)^{7}, we will use our answer from part (a) and get:

2x \times 590625x^{3} = 1181250x^{4}

To get the x^{4} term from -3(3x - 5)^{7}, we will use the general term of the Binomial Theorem formula to find the x^{4} in the expansion of (3x - 5)^{7}:

^{7}C_{3} \times (3x)^{7 - 3} \times (-5)^{3}

Using the calculator to work further we get:

35 \times 81x^{4} \times -125 = -354375x^{4}

Therefore the x^{4} term from -3(3x - 5)^{7} is:

(-3)(-354375x^{4}) = 1063125x^{4}

Hence the coefficient of the x^{4} term in the expansion of (2x - 3)(3x - 5)^{7} is:

1181250 + 1063125 =

\boldsymbol{{\color{Purple} = 2244375}}

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Finding expressions for p and for q:

p = ^{5}\textrm{C}_{1} \times (2x)^{4} \times k = 80kx^{4}

q = ^{5}\textrm{C}_{2} \times (2x)^{3} \times k^{2} = \frac{5!}{2! \times 3!} \times (2x)^{3} \times k^{2} =80k^{2}x^{3}

q - p = 80k^{2} - 80k = 480

k^{2} - k - 6 = 0\Rightarrow (k - 3)(k + 2) = 0

\boldsymbol{{\color{Purple} k = 3}} or \boldsymbol{{\color{Purple} k = - 2}}

.

Using the binomial theorem formula:

^{a^{2}-2}C_{2} \times (3x^{3})^{2} \times (-2)^{a^{2} - 4} = 3354624

Working out ^{a^{2}-2}C_{2}:

\frac{(a^{2}-2)!}{2!(a^{2}-4)!} = \frac{(a^{2} - 2)(a^{2} - 3)}{2}

So we get:

\frac{(a^{2} - 2)(a^{2} - 3)}{2} \times 9 \times (-2)^{a^{2}-4} = 3354624

Using the calculator to solve, we get that:

\boldsymbol{{\color{Purple} a = 4}}

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