Question

find the sum of each geometric series described.

1. first six terms of \\(3 + 9 + 27 + \cdots\\)
2. first nine terms of \\(0.5 + (-1) + 2 + \cdots\\)
3. first eight terms of \\(2 + 2\sqrt{3} + 6 + \cdots\\)
4. first \\(n\\) terms of \\(a_1 = 4, a_n = 2000, r = -3\\)
5. first \\(n\\) terms of \\(a_1 = 5, a_n = 1,310,720, r = 4\\)
6. first \\(n\\) terms of \\(a_1 = 3, a_n = 46,875, r = -5\\)
7. first \\(n\\) terms of \\(a_1 = -8, a_n = -256, r = 2\\)
8. first \\(n\\) terms of \\(a_1 = -36, a_n = 972, r = 7\\)

Explanation:

Identify series parameters for questions 40 to 42

Using the formula for the sum of a finite geometric series \(S_n = \frac{a_1(1-r^n)}{1-r}\):

• Question 40: \(a_1 = 3\), \(r = 3\), \(n = 6\).

$$S_6 = \frac{3(1-3^6)}{1-3} = \frac{3(1-729)}{-2} = \frac{3(-728)}{-2} = 1092$$

• Question 41: \(a_1 = 0.5\), \(r = -2\), \(n = 9\).

$$S_9 = \frac{0.5(1-(-2)^9)}{1-(-2)} = \frac{0.5(1+512)}{3} = \frac{256.5}{3} = 85.5$$

• Question 42: \(a_1 = 2\), \(r = \sqrt{3}\), \(n = 8\).

$$S_8 = \frac{2(1-(\sqrt{3})^8)}{1-\sqrt{3}} = \frac{2(1-81)}{1-\sqrt{3}} = \frac{-160}{1-\sqrt{3}} = \frac{160}{\sqrt{3}-1} = 80(\sqrt{3}+1)$$

Identify series parameters for questions 43 to 45

Using the alternative sum formula \(S_n = \frac{a_1 - a_n r}{1-r}\):

• Question 43: \(a_1 = 4\), \(a_n = 2000\), \(r = -3\).

$$S_n = \frac{4 - 2000(-3)}{1-(-3)} = \frac{4 + 6000}{4} = 1501$$

• Question 44: \(a_1 = 5\), \(a_n = 1310720\), \(r = 4\).

$$S_n = \frac{5 - 1310720(4)}{1-4} = \frac{5 - 5242880}{-3} = \frac{-5242875}{-3} = 1747625$$

• Question 45: \(a_1 = 3\), \(a_n = 46875\), \(r = -5\).

$$S_n = \frac{3 - 46875(-5)}{1-(-5)} = \frac{3 + 234375}{6} = \frac{234378}{6} = 39063$$

Identify series parameters for questions 46 to 47

Using the alternative sum formula \(S_n = \frac{a_1 - a_n r}{1-r}\):

• Question 46: \(a_1 = -8\), \(a_n = -256\), \(r = 2\).

$$S_n = \frac{-8 - (-256)(2)}{1-2} = \frac{-8 + 512}{-1} = -504$$

• Question 47: \(a_1 = -36\), \(a_n = 972\), \(r = -3\) (Note: The image shows \(r = -3\) or \(r = -7\), but \(a_n = a_1 r^{n-1} \implies 972 = -36 r^{n-1} \implies r^{n-1} = -27\), which yields \(r = -3\)).

$$S_n = \frac{-36 - 972(-3)}{1-(-3)} = \frac{-36 + 2916}{4} = \frac{2880}{4} = 720$$

Answer:

No.	Answer
41	\(85.5\)
42	\(80(\sqrt{3}+1)\) or \(80\sqrt{3}+80\)
43	\(1501\)
44	\(1747625\)
45	\(39063\)
46	\(-504\)
47	\(720\)