Special Sums Useful Series

7/16/2013 Special Sums 2 Notation Manipulation Consider sequences { a n } and constant c Factoring Summation Notation ∑ k=1 n akak c = c a 1 + c a 2 + c a 3 + … + c a n = c ( a 1 + a 2 + a 3 + … + a n ) ∑ k=1 n akak c =

7/16/2013 Special Sums 3 Notation Manipulation Consider sequences { a n } and constant c Summation Notation ∑ k=1 n akak c ∑ n akak c = ∑ 2k k=1 5 = ∑ k 5 2 = 2( ) = 30 Example:

7/16/2013 Special Sums 4 Notation Manipulation Consider sequences { a n }, { b n } and constant c Addition: Summation Notation ∑ k=1 n bkbk + ∑ n akak = ∑ n akak bkbk + ( ) ∑ 5 k 2 + ( ) = ∑ 5 k + ∑ 5 2 = 5(5 + 1) 2 + 5(2) = 25 Example:

7/16/2013 Special Sums 5 Sum of the First n Natural Numbers Consider Summation Notation ∑ k=1 n k = … + (n – 2) + (n – 1) + n Add OR =n + (n – 1) + (n – 2) + … ∑ k=1 n k ∑ n k2 = (n + 1) + (n + 1) + (n + 1) + … + (n + 1) giving

7/16/2013 Special Sums 6 Sum of the First n Natural Numbers resulting in Summation Notation ∑ k=1 n k2 = (n + 1) + (n + 1) + (n + 1) + … + (n + 1) = n(n + 1) k=1 n ∑ k = n(n + 1) 2 Thus n terms

7/16/2013 Special Sums 7 Sum of the First n Natural Numbers Summation Notation k=1 n ∑ k = n(n + 1) 2 Examples: ∑ k=1 100 k = 100( ) 2 5,050= ∑ k=1 495 k = 495( ) 2 122,760=

7/16/2013 Special Sums 8 Sum of the First n Natural Numbers – Another Way C Summation Notation ∑ k=1 n+1n+1 k2k2 = (n + 1) 2 + ∑ k=1 n k2k2 (k + 1) 2 ∑ k=0 n = (k 2 + 2k + 1) ∑ k=0 n = But also … Consider: ∑ k=1 n+1n+1 k2k2 + ∑ n k 2 + ∑ k=0 n 1 k2k2 ∑ k=1 n =

7/16/2013 Special Sums 9 Sum of the First n Natural Numbers – Another Way Summation Notation (n + 1) 2 + ∑ k=1 n k2k2 + ∑ n k 2 + ∑ k=0 n 1 k2k2 ∑ k=1 n = (n + 1) 2 ∑ k=1 n k 2 + ∑ k=0 n 1 = ∑ k=1 n k 2 =(n + 1) 2 – (n + 1) = ( (n + 1) – 1 ) = (n + 1)(n) (1)(1) WHY ?

7/16/2013 Special Sums 10 Sum of the First n Natural Numbers – Another Way Summation Notation ∑ k=1 n k 2 = (n + 1)(n) Thus ∑ k=1 n k = 2 n(n + 1) The Average of n and n 2

7/16/2013 Special Sums 11 Sum of the First n Natural Numbers – Another Way Note that we started with the sum of the squares of the first n natural numbers Could the sum of the first n cubes lead us to the sum of the first n squares? Let us find out … Summation Notation

7/16/2013 Special Sums 12 Sum of the First n Squares Co Summation Notation ∑ k=1 n+1n+1 k3k3 = (n + 1) 3 + ∑ k=1 n k3k3 n+1n+1 ∑ k3k3 ∑ k=0 n (k + 1) 3 = (k 3 + 3k 2 + 3k + 1) ∑ k=0 n =+ ∑ k=1 n k2k2 3 + ∑ k=0 n 1 k3k3 ∑ k=1 n = k ∑ n 3 + But also … Consider

7/16/2013 Special Sums 13 Sum of the First n Squares Summation Notation (n + 1) 3 + ∑ k=1 n k3k3 + ∑ n k2k2 3 + ∑ k=0 n 1 k3k3 ∑ k=1 n = k ∑ n 3 + (n + 1) 3 = ∑ k=1 n k2k2 3 + ∑ k=0 n 1 k ∑ k=1 n 3 + ∑ n k2k2 3 = (n + 1) 3 k ∑ k=1 n 3 – ∑ k=0 n 1 – = (n + 1) 3 – 3 ( n(n + 1) 2 ) – (n + 1)

7/16/2013 Special Sums 14 Sum of the First n Squares Summation Notation ∑ k=1 n k2k2 3 = (n + 1) 3 – 3 ( n(n + 1) 2 ) – (n + 1) = ( ) 2n 2 + 4n + 2 – 3n – 2 2 = n(n + 1)(2n + 1) 6 ∑ k=1 n k2k2

7/16/2013 Special Sums 15 Sum of the First n P th Powers Strategy: start with Write as Expand both sides and set equal Eliminate Solve for Summation Notation ∑ k=1 n+1n+1 k P+1 ∑ k=0 n (k + 1) P+1 (n + 1) P+1 ∑ k=1 n k P+1 + ∑ k=1 n k P+1 ∑ k=1 n kPkP and as

7/16/2013 Special Sums 16 Think about it !