Expand menu menu_open Minimize Start chapters Home History history History expand_more
{{ item.displayTitle }}
navigate_next
No history yet!
Progress & Statistics equalizer Progress expand_more
Student
navigate_next
Teacher
navigate_next
{{ filterOption.label }}
{{ item.displayTitle }}
{{ item.subject.displayTitle }}
arrow_forward
No results
{{ searchError }}
search
menu_open
{{ courseTrack.displayTitle }}
{{ statistics.percent }}% Sign in to view progress
{{ printedBook.courseTrack.name }} {{ printedBook.name }}
search Use offline Tools apps
Login account_circle menu_open

Successive Differences of Powers

Method

Successive Differences of Powers

For a sequence that is given by the explicit rule an=a1np, a_n=a_1\cdot n^p, where pp is an integer, a1a_1 and pp can be found by calculating the differences between consecutive terms of the sequence.
Consider the sequence

which is given by the rule above.

1

Subtract consecutive terms

First it is necessary to calculate the differences between the terms. Here there differences are as follows.

ana_n an1a_{n-1} anan1a_n-a_{n-1} Difference
208208 121.5121.5 288121.5288-121.5 166.5166.5
562.5562.5 288288 562.5288562.5-288 274.5274.5
972972 562.5562.5 972562.5972-562.5 409.5409.5
1543.31543.3 972972 1543.59721543.5-972 571.5571.5


2

Create a new sequence

A new sequence can then be created using these differences.


3

Repeat until the differences are constant


4

Identify pp

The integer pp is the same as the number of differences needed. Here, since the 3rd3^{\text{rd}} the differences were calculated gave identical results pp equals 3.3.

5

Calculate a1a_1

a1a_1 can be calculated by using the relationship d=a1p(p1)1, d=a_1\cdot p \cdot (p-1) \cdot \ldots \cdot 1, where dd is the number on row number p.p.
Here, we have d=27d=27 and p=3.p=3. 27=a1321a1=4.5 27=a_1\cdot 3 \cdot 2 \cdot \cdot 1 \quad \Leftrightarrow \quad a_1=4.5