Summation Properties : Discrete Math : Lecture

Lecture#15 Discrete Mathematics Summation The sum of the terms of a sequence forms a series. If a1, a2, a3, … represent a sequence of numbers, then the corresponding series is a1 + a2 + a3 + … = The Greek letter sigma is used to write a sum in short hand notation = a1 + a2 + a3 + … + an where k varies from 1 to n represents the sum. Computing Summation Let a0 = 2, a1 = 3, a2 = -2, a3 = 1 and a4 = 0. Compute each of the summations: = a0 + a1 + a2 +a3 +a4 = 2 + 3 + -2 + 1 + 0 = 4 = a0 + a2 +a4 = 2 + (-2) + 0 = 0 = a1 = 3 Computing Summation: Examples Compute each of the following summations: Expanding Summation Write the summation to expanded form: Expanded Form of Summation Write the following using summation notation: Summation and Variables Consider = 1 + 4 + 9 = 14 & = 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. Properties of Summation c?R Properties of Summation n terms Summation Properties: Exercise Express the following summation more simply: 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 Let a be the first term and d be the common difference of an arithmetic series. Then its nth term is: an = a + (n - 1)d; n ? 1 If Sn denotes the sum of first n terms of the A.S, then Sn = a + (a + d) + (a + 2d) + … + [a + (n-1) d] Sn = a + (a+d) + (a + 2d) + … + an Sn = a + (a+d) + (a + 2d) + … + (an - d) + an ………(1) Where an = a + (n - 1) d Rewriting the terms in the series in reverse order, Sn = an + (an - d) + (an - 2d) + … + (a + d) + a ……….(2) Adding (1) & 2, we get Sn= n/2 [2 a + (n - 1) d] Sum of (N-Term) Arithmetic Series Find the sum of first n natural numbers: Solution: A = 1, d = 2-1=1, n = n Sum of Arithmetic Series: Example Find the sum of all two digit positive integers which are neither divisible by 5 nor by 2. Solution: Sn =(11 + 13 + 15 + …. +99)–(15 + 25 + 35 +…..+ 95) Sum of Arithmetic Series: Example 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 + ar2 + ar3 + … Geometric Series 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 Sn denotes the sum of first n terms of the G.S. then Sn = a + ar + ar2 + ar3 + … + arn-2 + arn-1……………(1) Multiplying both sides by r we get. r Sn = ar + ar2 + ar3 + … + arn-1 + arn………………(2) Subtracting (2) from (1) we get Sn - r Sn = a – arn ? (1 - r) Sn = a (1 - rn) ? Sn = a (1 - rn) / (1 - r) (r <>1) Sum of Geometric Series (N-Terms) Find the sum of geometric series given below: Geometric Series: Example Consider the infinite geometric series a + ar + ar2 + … + arn-1 + … then Infinite Geometric Series If Sn ? S as n ? ?, then the series is convergent and S is its sum. If |r| < 1, then rn ? 0 as n ? ? Find the sum of the infinite geometric series: Geometric Series Sum: Example 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 Important Sums Series Sum: Example Sum to n terms the series 1.5+5.11+9.17+… Solution: Lets write the Nth term of given series Tk = [A0 + (k-1)d].[B0 + (k-1)d] = [1+(k-1)4].[5+(k-1)6] = (4k-3).(6k-1) = 24k2-22k+3 Now Sn = Series Sum: Example = (24k2-22k+3) = 24 k2 – 22 k + 3 1 = 24 (n(n+1)(2n+1))/6 – 22 (n(n+1)) / 2 + 3 n = 4n (2n2 + 3n + 1) - 11 (n2 + n) + 3n = 8n3 + 12n2 + 4n - 11n2 - 11n + 3n = 8n3 + n2 - 4n