1. Lecture#15 Discrete Mathematics

2. Summation

3. Computing Summation Let a 0 = 2, a 1 = 3, a 2 = -2, a 3 = 1 and a 4 = 0. Compute each of the summations: = a 0 + a 1 + a 2 +a 3 + a 4 = 2 + 3 + -2 + 1 + 0 = 4 = a 0 + a 2 + a 4 = 2 + (-2) + 0 = 0 = a 1 = 3

4. Computing Summation: Examples Compute each of the following summations:

5. Expanding Summation Write the summation to expanded form:

6. Expanded Form of Summation Write the following using summation notation:

7. Summation and Variables Consider = 1 + 4 + 9 = 14 & hence  The index of a summation can be replaced by any other symbol. The index of a summation is therefore called a dummy variable.

8. Properties of Summation cRcR

9. n terms

10. Summation Properties: Exercise Express the following summation more simply:

11. The sum of the terms of an arithmetic sequence forms an arithmetic series (A.S). For example 1 + 3 + 5 + 7 + … is an arithmetic series of positive odd integers. In general, if a is the first term and d the common difference of an arithmetic series, then the series is given as: a + (a+d) + (a+2d) +… Arithmetic Series

12. Let a be the first term and d be the common difference of an arithmetic series. Then its nth term is: a n = a + (n - 1)d;n  1 If S n denotes the sum of first n terms of the A.S, then S n = a + (a + d) + (a + 2d) + … + [a + (n-1) d] S n = a + (a+d) + (a + 2d) + … + a n S n = a + (a+d) + (a + 2d) + … + (a n - d) + a n ………(1) Where a n = a + (n - 1) d Rewriting the terms in the series in reverse order, S n = a n + (a n - d) + (a n - 2d) + … + (a + d) + a ……….(2) Adding (1) & 2, we get S n = n/2 [2 a + (n - 1) d] Sum of (N-Term) Arithmetic Series

13. Find the sum of first n natural numbers: Solution: A = 1, d = 2-1=1, n = n Sum of Arithmetic Series: Example

14. Find the sum of all two digit positive integers which are neither divisible by 5 nor by 2. Solution: S n =(11 + 13 + 15 + …. +99)–(15 + 25 + 35 +…..+ 95) Sum of Arithmetic Series: Example

15. The sum of the terms of a geometric sequence forms a geometric series (G.S.). For example 1 + 2 + 4 + 8 + 16 + … is geometric series. In general, if a is the first term and r the common ratio of a geometric series, then the series is given as: a + ar + ar 2 + ar 3 + … Geometric Series

16. Let a be the first term and r be the common ratio of a geometric series. Then its nth term is: an = arn-1;n  1 If S n denotes the sum of first n terms of the G.S. then S n = a + ar + ar 2 + ar 3 + … + ar n-2 + ar n-1 ……………(1) Multiplying both sides by r we get. r S n = ar + ar 2 + ar 3 + … + ar n-1 + ar n ………………(2) Subtracting (2) from (1) we get S n - r S n = a – ar n  (1 - r) S n = a (1 - r n )  S n = a (1 - r n ) / (1 - r) (r <>1) Sum of Geometric Series (N-Terms)

17. Find the sum of geometric series given below: Geometric Series: Example

18. Consider the infinite geometric series a + ar + ar2 + … + arn-1 + … then Infinite Geometric Series If S n  S asn  , then the series is convergent and S is its sum. If |r| < 1, then r n  0 as n  

19. Find the sum of the infinite geometric series: Geometric Series Sum: Example

20. Find a common fraction for the recurring decimal 0.81 Solution: 0.81= 0.8181818181 … = 0.81 + 0.0081 + 0.000081 + … which is an infinite geometric series with Geometric Series Sum: Example

21. Important Sums

22. Series Sum: Example Sum to n terms the series 1.5+5.11+9.17+… Solution: Lets write the Nth term of given series T k = [A 0 + (k-1)d].[B 0 + (k-1)d] = [1+(k-1)4].[5+(k-1)6] = (4k-3).(6k-1) = 24k 2 -22k+3 Now S n =

23. Series Sum: Example = (24k 2 -22k+3) = 24 k 2 – 22 k + 3 1 = 24 (n(n+1)(2n+1))/6 – 22 (n(n+1)) / 2 + 3 n = 4n (2n 2 + 3n + 1) - 11 (n 2 + n) + 3n = 8n 3 + 12n 2 + 4n - 11n 2 - 11n + 3n = 8n 3 + n 2 - 4n