Sequences and Series (Unit 1C) Explicit Formula & Summing a Series

Arithmetic Sequence Example: 6, 8, 10, 12,… An arithmetic sequence is a sequence of numbers that has a constant difference between every two consecutive terms. Example: 6, 8, 10, 12,… Each term is increasing by 2 Constant difference just means each successive number increases or decreases by the same amount!

Arithmetic Series An arithmetic series is the sum of the terms in an arithmetic sequence. Ex. 1, 2, 3, 4… (Arithmetic sequence) 1 + 2 + 3 + 4… (Arithmetic series)

Arithmetic Sequence To find an exact term in an arithmetic sequence use the ________ formula. Explicit an = d(n – 1) + a1 Insert the common difference Insert the 1st term

Arithmetic Sequence Find the explicit formula for the following sequence: 5, 11, 17, 23, ... 6 6 6 an = d(n – 1) + a1 5 6 Insert the common difference Insert the 1st term

Try Again: Find the explicit formula for the following sequence: 15, 13, 11, 9 ... -2 -2 -2 an = d(n – 1) + a1 = - 83 -2 15 50 50 Now type -2 (50 - 1) +15 = in your calculator The 50th term in this sequence is -83 Insert the common difference Insert the 1st term Now, suppose I want to know the 50th term in this sequence. Use the Explicit Formula above and let n = 50.

Try Again: Find the explicit formula for the following sequence: 50, 45.5, 41, 36.5 ... When difficult, Use your calculator to find the difference by subtracting 2nd term from the 1st term: 45.5 - 50.0_ -4.5 = d -4.5 -4.5 -4.5 an = d (n – 1) + a1 -4.5 50  Using your calculator, Input what you see that comes after the = sign a25 = -4.5 (25 – 1) + 50 Insert the common difference a25 = -58 is the 25th term. Insert the 1st term Find the 25th term in this sequence. Use the Explicit Formula above and replace n with 25.

The can pyramid… How many cans are there in this pyramid? How many cans are there in a pyramid with 100 cans on the bottom row?

Back to the pyramid… We wanted to work out the sum of: 1 + 2 + 3 + ….. + 98 + 99 + 100 If we write it out in reverse we get…. 100 + 99 + 98 + ….. + 3 + 2 + 1 101 + 101 + 101 +….. 101 x 100 = 10100 How many times do we add 101 together? 10100 / 2 = 5050 That’s correct! Half our answer or simply divide by 2! The summing pattern above is cool, but our answer is now TOO big. What do we need to do to get the correct answer?

Let’s Review Steps We wanted to work out the sum of: 1 + 2 + 3 + ….. + 98 + 99 + 100 100 + 99 + 98 + ….. + 3 + 2 + 1 101 + 101 + 101 +….. 101 x 100 pairs = 10100 10100 / 2 = 5050 That’s correct! Half our answer or simply divide by 2!

Activity 1 Work out the sum of the first 50 positive integers (1+2+3+...+50) using the same method as used with the pyramid. Stop Here... And allow students time to work this problem on their paper before moving ahead.

Activity 1 Work out the sum of the first 50 positive integers (1+2+3+...+50) using the same method as the pyramid. Check your answer.... Did you get 1275? If so you are CORRECT!

Activity 2 Work out the sum of all the ODD numbers from 21 up to 99 using the same method as before. You MAY need to think a bit harder to determine the total number of pairs since we did not start at 1! Remember to write your list of numbers you are summing forwards and backwards just as you were shown earlier with the pyramid problem! You CAN DO IT ! STOP HERE! Allow students time to work the problem before moving ahead!

Find the Sum of all the odd numbers from Arithmetic Series Find the Sum of all the odd numbers from 21 up to 99 ..... SOLUTION! 21 + 23 + 25 + … + 95 + 97 + 99 99 + 97 + 96 + … + 25 + 23 + 21 __________________________________________________________ 120 +120 + 120 + … +120 + 120 + 120 Did you get there are 40 pairs of numbers? n = 40 SUM = 40(120) = 4800 divided by 2 = 2400 Because we counted every number twice The SUM of this sequence, S40 = 2400!

Arithmetic Series Wah Lah! 21 + 23 + 25 + … + 95 + 97 + 99 a1 = first term an = last term 21 + 23 + 25 + … + 95 + 97 + 99 99 + 97 + 96 + … + 25 + 23 + 21 __________________________________________________________ 120 +120 +120 + … +120 + 120 + 120 We had 40(120) = 4800 divided by 2 = 2400. You have uncovered a FORMULA for summing series! Sum n terms = nbr. of terms (first term + last term) divided by 2 Sn = n (a1 + an) 2 S40 = 40 (21 + 99)  40 (120)  2400 2 2 Wah Lah!

Student Practice Find the sum of the following series using the formula 1) Given: 1+ 2+ 3 + 4 +… + 999 + 1000 S1000 = _____ 2) Given: 2 +4 + 6 + 8 +… + 598 + 600 S600 = _____ Sn = n (a1 + an) 2 STOP HERE! Allow students time to work the problems before moving ahead!

Student Practice Find the sum of the following series using the formula 1) Given: 1+ 2+ 3 + 4 +… + 999 + 1000 S1000 = _____ 2) Given: 2 +4 + 6 + 8 +… + 598 + 600 S600 = _____ Sn = n (a1 + an) 2 S1000 = 1000 (1 + 1000) = 500(1001) = 500,500 2 S600 = 600 (2 + 600) = 300(602) = 180,600 2

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