Sequences, Series, and Induction: math test solutions, Grade 12 – download the PDF

1 Arithmetic sequence / 3 pts

(a) \(5d = 31 - 11 = 20\), so \(d = 4\). (1 pt)

(b) \(a_1 = 11 - 2\times 4 = 3\) and \(a_n = 3 + 4(n-1) = 4n - 1\). (1 pt)

(c) \(a_{25} = 99\), so \(S_{25} = \dfrac{25(3 + 99)}{2} = 25 \times 51 = 1275\). (1 pt)

2 Geometric sequence / 3 pts

\(a_5 / a_2 = r^3 = 27\), so \(r = 3\). (1 pt)

\(a_1 = 12 / 3 = 4\). (1 pt)

\(S_7 = 4\cdot\dfrac{3^7 - 1}{3 - 1} = 2 \times 2186 = 4372\). (1 pt)

3 Infinite series / 3 pts

(a) \(a_1 = 20\), \(r = -\dfrac12\), \(|r| \lt 1\). \(S = \dfrac{20}{3/2} = \dfrac{40}{3}\). (2 pts)

(b) \(0.27 + 0.0027 + \cdots\): \(a_1 = 0.27\), \(r = 0.01\), so \(S = \dfrac{0.27}{0.99} = \dfrac{27}{99} = \dfrac{3}{11}\). (1 pt)

4 Sigma notation and recursion / 3 pts

(a) \(2 + 4 + 8 + 16 + 32 + 64 = 126\). (1 pt)

(b) \(b_2 = 2 + 4 = 6\), \(b_3 = 6 + 6 = 12\), \(b_4 = 12 + 8 = 20\). (1 pt)

The terms 2, 6, 12, 20 equal \(1\cdot 2,\ 2\cdot 3,\ 3\cdot 4,\ 4\cdot 5\), so \(b_n = n(n+1)\). (1 pt)

5 Proof by induction / 4 pts

Base case. \(n = 1\): left side 1, right side \(\dfrac{3 - 1}{2} = 1\). (1 pt)

Inductive hypothesis. Assume \(1 + 3 + \cdots + 3^{k-1} = \dfrac{3^k - 1}{2}\). (1 pt)

Inductive step. Adding \(3^k\): \(\dfrac{3^k - 1}{2} + 3^k = \dfrac{3^k - 1 + 2\cdot 3^k}{2} = \dfrac{3^{k+1} - 1}{2}\). (1 pt)

Conclusion. By induction the formula is true for all \(n \ge 1\). (1 pt)

6 Binomial Theorem / 4 pts

(a) The coefficients are 1, 5, 10, 10, 5, 1 and the powers of \(-2\) are \(1, -2, 4, -8, 16, -32\):

\((x-2)^5 = x^5 - 10x^4 + 40x^3 - 80x^2 + 80x - 32\). (3 pts)

(b) \(\dbinom{7}{3}\cdot 2^3 = 35 \times 8 = 280\). (1 pt)