Math AA SL

(a) (i)

\boldsymbol{{\color{Purple} \textbf{Sequence 4}}}

(a) (ii)

\boldsymbol{{\color{Purple} \textbf{Sequence 1}}}

(b) (i)

d = 2 - 6

\boldsymbol{{\color{Purple} d = -4}}

(b) (ii)

Using the nthe term of an arithmetic sequence formula:

u_{15} = 6 + (15-1)(-4)

\boldsymbol{{\color{Purple} u_{15} = -50}}

(c) (i)

\boldsymbol{{\color{Purple} r = \frac{1}{3}}}

(c) (ii)

Using the sum of nterms of a finite geometric sequence formula:

S_{7} = \frac{9\left ( 1 - \left (  \frac{1}{3}\right )^{7} \right )}{1 - \frac{1}{3}}

\boldsymbol{{\color{Purple} S_{7} = \frac{1093}{81}}}

.

(a)

r = \frac{112}{-280}

\boldsymbol{{\color{Purple} r = -0.4}}

(b)

Using the nthe term of a geometric sequence formula:

u_{12} = -280 \times \left ( -0.4 \right )^{12-1}

\boldsymbol{{\color{Purple} u_{12} \approx 0.0117}}

(c)

Using the sum of an infinite geometric sequence formula:

S_{\infty } = \frac{-280}{1-(-0.4)}

\boldsymbol{{\color{Purple} S_{\infty } = -200}}

.

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}}

.

(a)

Using the compound interest formula (you can also use the Finance app on your calculator):

6000\left ( 1 + \frac{8.5}{100 \times 2} \right )^{2 \times 7}\approx

\boldsymbol{{\color{Purple} \approx 10745}}

(b)

Using the compound interest formula (you can also use the Finance app on your calculator):

13000 = 6000\left ( 1 + \frac{8.5}{100 \times 2} \right )^{2 \times n}

Using the calculator to solve for n:

n = 9.288, therefore

\boldsymbol{{\color{Purple} n = 10}}

(c)

Using the compound interest formula (you can also use the Finance app on your calculator):

14000 = 6000\left ( 1 + \frac{r}{100 \times 4} \right )^{4 \times 11}

Using the calculator to solve for r we get that:

\boldsymbol{{\color{Purple} r \approx 7.78}}

(d)

Adding 70\: \% of what was added the previous year means that the amounts added each year form a geometric sequence where r = 0.7.

Using the sum of an infinite geometric sequence formula:

S_{\infty} = \frac{4500}{1 - 0.7} = 15000

\boldsymbol{{\color{Purple} \textbf{Since}\:  15000 < 17500, \textbf{Alma will not reach her goal.}}}

(e)

Using the sum of n terms of a geometric sequence formula:

17500 = \frac{4500(1 - r^{20})}{1 - r}

Using the calculator to solve for r:

r \approx 0.743543, so:

\boldsymbol{{\color{Purple} x \approx 74.4}}

.

(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}}

.

(a)

Using the compound interest formula (you can also use the TVM Solver inside the Finance app), we get:

FV = 3500\left ( 1 + \frac{1.87}{100(12)} \right )^{(12)(8)} \approx

\boldsymbol{{\color{Purple} \approx \$4064.32}}

(b)

4064.32 - 3500 =

\boldsymbol{{\color{Purple} = \$ 564.32}}

(c)

Using the TVM Solver inside the Finance app (you can also use the compound interest formula), we get:

N = 8 \times 4 = 32

I\, \% = 2.05

FV = 4064.32

P/Y = 4

C/Y = 4

Solving for PV we get that:

\boldsymbol{{\color{Purple} P \approx 3451.00}}

.

(a)

Using the nth term of an arithmetic sequence formula we get:

u_{n} = -86 + (n-1) \times 4

Substituting zero for u_{n} and solving for n we get:

0 = -86 + (n-1) \times 4

n = 22.5

Since n must be a whole number, the final answer is:

\boldsymbol{{\color{Purple} n = 23}}

(b)

Using the sum of the first n terms of a arithmetic sequence formula we get:

S_{n} = \frac{n}{2}\left ( 2(-86) + (n-1) \times 4 \right )

Finding the minimum value of this expression, for example, by graphing on the calculator, we get that the minimum value of S_{n} is:

\boldsymbol{{\color{Purple} -968}}

.

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)

\boldsymbol{{\color{Purple} r = 0.7}}

(b)

u_{1} = S_{1}= \sum_{k = 1}^{1}3 \times \left ( 0.7 \right )^{k} = 3 \times (0.7)^{1}

 \boldsymbol{{\color{Purple} u_{1} = 2.1}}

(c)

Using the sum of n terms of a finite geometric sequence formula:

\frac{2.1(1 - 0.7^{n})}{1-0.7}<6.999

Using the calculator to solve (graphing, Numeric Solver, or table):

S_{24} = 6.9986589

S_{25} = 6.9990613 

Therefore:

\boldsymbol{{\color{Purple} n = 24}}

.

(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}}

.

Using the sum of an infinite geometric sequence formula and we can write that:

100 = \frac{u_{1}}{1 - r}

Using the sum of n terms of a finite geometric sequence we can write that:

67.232 = \frac{u_{1}(1 - r^{5})}{1 - r}

Rearranging the first equation we get that:

u_{1} = 100(1 - r)

Substituting 100(1 - r) for u_{1} in the second equation we can write that:

67.232 = \frac{(100(1 - r))(1 - r^{5})}{1 - r}

Simplifying gives us:

67.232 = 100(1 - r^{5})

Solving for r either by graphing or using the Solver function on the calculator, we get  that:

\boldsymbol{{\color{Purple} r = 0.8}}

.

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}}

.

(a)

We can calculate a 7\, \% decrease by multiplying the initial number of raspberry bushes by 0.93.

So for the number of raspberry bushes at the end of 2024 we get:

700 \times 0.93 + 80

For the number of raspberry bushes at the end of 2025 we get:

(700 \times 0.93 + 80) \times 0.93 + 80 = 700 \times 0.93^{2} + 80 \times 0.93 + 80 = 700 \times 0.93^{2} + 80(0.93 + 1)

For the number of raspberry bushes at the end of 2026 we get:

(700 \times 0.93^{2} + 80(0.93 + 1)) \times 0.93 + 80 = 700 \times 0.93^{3} + 80(0.93^{2} + 0.93 + 1)

Following this pattern, for the number of raspberry bushes at the end of 2030 we get:

700 \times 0.93^{7} + 80(0.93^{6} + 0.93^{5} + 0.93^{4} + 0.93^{3} + 0.93^{2} + 0.93 + 1) \approx

\boldsymbol{{\color{Purple} \approx 876 \: \textbf{bushes}}}

(b)

Let’s write an expression for the number of raspberry bushes n years after 2023:

700 \times 0.93^{n} + 80(0.93^{n-1} + 0.93^{n-2} + ... + 0.93 + 1)

The expression in the brackets is the sum of the first n terms of a geometric sequence with u_{1} = 1 and r = 0.93, so for the number of raspberry bushes n years after 2023 we get:

700 \times 0.93^{n} + 80\left ( \frac{1(1 - 0.93^{n})}{1 - 0.93} \right )

We are looking for the smallest value of n for which the number of raspberry bushes exceeds 1100, so simplifying and equating the expression to 1100 we get:

700 \times 0.93^{n} +  \frac{8000(1 - 0.93^{n})}{7} = 1100

Solving for n either by graphing or using the Solver function on the calculator, we get that n \approx 32.2, therefore the first year in which the number of raspberry bushes exceeds 1100 is 2023 + 33, so:

\boldsymbol{{\color{Purple} 2056}}

.