Fill in the table so it represents a linear function.

Mathematics

1 answer:

olga2289 [7]3 years ago

6 0

Answer:

(From left to right) 2 5 8

Step-by-step explanation:

1. Find the gap (between the first and last terms) 11 - (-1) = 12

2. Divide by the number of times the linear term is added: 12/4=3

3. Use the common difference to find the missing values.

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3 years ago

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3/2 x 4/3 x 5/4… x 2006/2005

Lady_Fox [76]

Answer:

1003

Step-by-step explanation:

The problem is a classic example of a telescoping series of products, a series in which each term is represented in a certain form such that the multiplication of most of the terms results in a massive cancelation of subsequent terms within the numerators and denominators of the series.

The simplest form of a telescoping product $a_{k} \ = \ \displaystyle\frac{t_{k}}{t_{k+1}}$ , in which the products of <em>n</em> terms is

$a_{1} \ \times \ a_{2} \ \times \ a_{3} \ \times \ \cdots \times \ a_{n-1} \ \times \ a_{n} \ = \ \displaystyle\frac{t_{1}}{t_{2}} \ \times \ \displaystyle\frac{t_{2}}{t_{3}} \ \times \ \displaystyle\frac{t_{3}}{t_{4}} \ \times \ \cdots \ \times \ \displaystyle\frac{t_{n-1}}{t_{n}} \ \times \ \displaystyle\frac{t_{n}}{t_{n+1}} \\ \\ \-\hspace{5.55cm} = \ \displaystyle\frac{t_{1}}{t_{n+1}}.$ .

In this particular case, $t_{1} \ = \ 2$ , $t_{2} \ = \ 3$ , $t_{3} \ = \ 4$ , ..... , in which each term follows a recursive formula of $t_{n+1} \ = \ t_{n} \ + \ 1$ . Therefore,

$\displaystyle\frac{t_{2}}{t_{1}} \times \displaystyle\frac{t_{3}}{t_{2}} \times \displaystyle\frac{t_{4}}{t_{3}} \times \cdots \times \displaystyle\frac{t_{n}}{t_{n-1}} \times \displaystyle\frac{t_{n+1}}{t_{n}} \ = \ \displaystyle\frac{3}{2} \times \displaystyle\frac{4}{3} \times \displaystyle\frac{5}{4} \times \cdots \times \displaystyle\frac{2005}{2004} \times \displaystyle\frac{2006}{2005} \\ \\ \-\hspace{5.95cm} = \ \displaystyle\frac{2006}{2} \\ \\ \-\hspace{5.95cm} = 1003$