Sequence and Series Review Problems

Sequence and Series Review Problems

Sequence Review Problems
Example 1: Find the explicit formula for the following arithmetic sequence: The explicit formula for arithmetic sequences is: You must replace the a1 and d 1, 6, 11, 16, … an = a1 + (n – 1)d

First you must find the common difference between the numbers: 1, 6, 11, 16, … 5 5 d =

Then find a1 a1 1, 6, 11, 16, …

Plug them both into the equation: Distribute the d Combine like terms a1 = 1 an = 1 + (n – 1)5 d = 5 an = 1 + 5n – 5 an = 5n – 4

Practice 1 Put the following sequences into explicit form:
1, 3, 5, 7, … -2, 1, 4, 7, … 9, 6, 3, 0, … -67, -60, -53, …

Example 2: Find the recursive formula for the following arithmetic sequence: The recursive formula for an arithmetic sequence is: You need to find the difference a1 = 3, a5 = 15 an = an-1 + d

The difference formula: an = the second or last number in the sequence. It can be any number after the first number a1 represents the first number you’re given (if not a1, then use the earliest term)

In this case we have a1 and a5, so a1 will be the first term and a5 will be the last a1 = 3, a5 = 15

Plug d into the equation: an = an-1 + d an = an-1 + 3

Practice 2 Find the recursive formula for the following arithmetic sequences a1 = 2, a6 = 12 a1 = 17, a8 = -25 a3 = 5, a10 = 54 a7 = 23, a19 = -85

Example 3: Find the explicit formula for the following geometric sequence: The explicit formula for a geometric sequence is: We must replace the a1 and r 1, 2, 4, 8, …

r is the common ration determined by dividing the second from the first number: 1, 2, 4, 8, …

And you know a1 is the first term in the series a1 1, 2, 4, 8, …

Plug them both into the equation: DO NOT combine the numbers!!! In this case 1 times a number is just that numbers so you don’t need the 1 a1 = 1 r = 2

Practice 3 Find the explicit formula for the following geometric sequences 1, 3, 9, … 3, 12, 48, … 2, -10, 50, … 36, 18, 9, …

Example 4: Find the recursive formula for the following geometric sequence: The recursive formula for a geometric sequence is 27, 9, 3, … an = r*an-1 times

Find r by dividing the second term by the first 27, 9, 3, …

Plug r into the equation an = r*an-1

Practice 4 Find the recursive formulas for the following geometric sequences 6, 12, 24, … 180, 60, 20, … 8, -32, 128, … -1, -2, -4, …

Series Review Problems
Example 5: Find the sum of the arithmetic series This is an arithmetic series (any time you see the word SUM you’re using series) Since you have the first and last term, use the formula a1 = 3, a6 = 19

The sum of the first n terms The number of terms you’re adding

The last term you’re adding The first term you’re adding

Since you have the first and the 6th term, there are a total of 6 terms you’re adding so a1 = 3, a6 = 19 n = 6 Plug everything into the equation

* Simplify That is the sum of the first six numbers in the series

Example 6: Find the sum of the following series: In this case we don’t have a1, so we’ll use a5 as the starting point a5 = 5, a30 = 78

We need to find the number of terms we’re adding up We are starting with the 5th term, so we are leaving out terms 1, 2, 3, and 4 That means we are adding 30 – 4 terms So: n = 26

In the formula, use a5 for the first number and a30 for the last Use your calculator a5 = 5, a30 = 78 n = 26

Practice 5 Find the sum of the following arithmetic series
a1 = 2, a6 = 12 a1 = 17, a8 = -25 a3 = 5, a10 = 54 a7 = 23, a19 = -85

Example 7: Find the 40th partial sum of the following arithmetic series: We aren’t given the last term in the series, so we can use the other arithmetic series formula: …

40th partial sum means add the first 40 terms so: We can also find d: n = 40 … 3 3 d =

Plug everything into your formula

Practice 6 Find the 20th partial sum of the series 1 + 5 + 9 + …

Example 8: Find the sum of the geometric series: Use the geometric series formula: a1 = 2, r = 3, n = 6

We have everything we need to plug in and solve the problem a1 = 2 r = 3 n = 6

Practice 7 Find the sum of the following geometric series
a1 = 2, r = 4, n = 5 a1 = 3, r = -2, n = 10 a1 = 12, r = ½ , n = 6

Put the following series in sigma notation: There are 5 terms so the bottom is n = 1 and the top is 5 a1 a5

We now need an explicit formula to go in front The series is arithmetic since there is a common difference ? 3 3 d =

Use the arithmetic explicit formula: an = a1 + (n – 1)d an = 1 + (n – 1)3 an = 1 + 3n – 3 an = 3n – 2

Plug your formula into sigma 3n – 2

Practice 8 Put the following series into sigma notation

Solve: n = 6 Start at the first term Add to the sixth term

Since the formula is arithmetic (the n is not an exponent) we’ll use the formula: We need to find the first and last term (6th)

To find the first term, plug 1 into the equation Now plug it all into your formula a1 =3(1) + 2 = 5 a6 =3(6) + 2 = 20 n = 6

Solve

Solve: Start at the 3rd term Add to the 8th term

This is a geometric series since n is an exponent Use the formula: We need a1, r, and n

Plug in 3 to find the first term we’re adding r is the number being raised to the power n is the number we are adding up: 8 – 2 = 6 (remember, we’re leaving out the first 2 numbers)

Plug everything into the formula a3 = 4 r = 2 n = 6

Practice 9 Solve the following

Sequence and Series Review Problems

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