Question

Determine a formula for 1/1.2 + 1/2.3 + ... + 1/n.(n+1) (Enter the fraction in the form a/b.) For n=1, 1/1.2 + 1/2.3+....+ 1/n.1(n+1)=(Click to select) For n=2, 1/1.2 + 1/2.3+....+ 1/n.1(n+1)=(Click to select) For n=3, 1/1.2 + 1/2.3+....+ 1/n.1(n+1)=(Click to select) For n=4, 1/1.2 + 1/2.3+....+ 1/n.1(n+1)=(Click to select) For n=5, 1/1.2 + 1/2.3+....+ 1/n.1(n+1)=(Click to select) Identify the formula for the given sum, derived from the values obtained n=1 to 5. a. n/(n+1) b. n/(n-1) c. (n-1)/(n+1) d. (n+5)/n e. (n+2)(n+9)

Answer 1

The sum of the series 1/1.2 + 1/2.3 + ... + 1/n(n+1) is a telescoping series, which simplifies to n/(n+1), corresponding to option a. This is found by expressing each term as a difference and observing the cancellation of terms.

Step-by-step explanation:

The series in question is a telescoping series, which means that most terms cancel each other out when we expand the series. To determine a formula for this sum, note that each term can be expressed as a difference of fractions by partial fraction decomposition. For example, 1/n(n+1) can be written as 1/n - 1/(n+1). When we apply this to each term in the series:

• 1/1.2 becomes 1/1 - 1/2,
• 1/2.3 becomes 1/2 - 1/3,
• and so on, until
• 1/n(n+1) becomes 1/n - 1/(n+1).

When you expand the series with these terms, you will see that all the terms cancel out except for the very first 1/1 and the very last -1/(n+1). This means that the sum of the series from 1/1.2 + 1/2.3 + ... + 1/n(n+1) is simply 1 - 1/(n+1).

Simplifying, the formula becomes 1 - 1/(n+1) = (n+1-1)/(n+1) = n/(n+1). Therefore, the formula for the given sum is n/(n+1). This corresponds to option a. n/(n+1) from the given choices.