Question

2.2 inductive reasoning
name
find the next two terms of the sequences then write a conjecture.

1. 1, -2, 3, -4, 5, _, _ conjecture: ___
2. 0, 2, 6, 12, 20, _, _ conjecture: ___

3.

1. z, y, x, w, v, _, _, conjecture: ___
2. $\frac{3}{4}$, $\frac{3}{8}$, $\frac{3}{16}$, $\frac{3}{32}$, _, _ conjecture: ___

Explanation:

Step1: Analyze sequence 1 pattern

The sequence alternates sign and increases by 1 in absolute - value. The next two terms are - 6, 7. The conjecture is that the \(n\)th term is \(a_{n}=(-1)^{n + 1}n\).

Step2: Analyze sequence 2 pattern

The differences between consecutive terms are 2, 4, 6, 8. The next differences are 10 and 12. So the next terms are \(20+10 = 30\), \(30 + 12=42\). The conjecture is \(a_{n}=n(n - 1)\).

Step3: Analyze sequence 3 pattern

The number of sides of the polygons increases by 1. The next two are a hexagon and a heptagon. The conjecture is that the \(n\)th figure is a polygon with \(n + 2\) sides.

Step4: Analyze sequence 4 pattern

The letters are in reverse - alphabetical order. The next two are U, T. The conjecture is that the \(n\)th letter is the \((27 - n)\)th letter of the alphabet.

Step5: Analyze sequence 5 pattern

The numerator is 3 and the denominator doubles each time. The next two terms are \(\frac{3}{64},\frac{3}{128}\). The conjecture is \(a_{n}=\frac{3}{2^{n + 1}}\).

Answer:

1. - 6, 7; \(a_{n}=(-1)^{n + 1}n\)
2. 30, 42; \(a_{n}=n(n - 1)\)
3. Hexagon, Heptagon; The \(n\)th figure is a polygon with \(n + 2\) sides
4. U, T; The \(n\)th letter is the \((27 - n)\)th letter of the alphabet
5. \(\frac{3}{64},\frac{3}{128}\); \(a_{n}=\frac{3}{2^{n + 1}}\)