Investigating a Sequence of Numbers Essay Example

Investigating a Sequence of Numbers Essay Example Investigating a Sequence of Numbers Essay Investigating a Sequence of Numbers Essay In this Mathematics Portfolio, I am going to investigate a sequence of numbers by mathematical methods which I have learnt in the I.B. Mathematics HL course. Throughout the investigation, I will include all my workings in order to let examiners know exactly how I come up with the answers. A sequence is a set of numbers with a definite order. A series is a sum of a sequence. The sequence of numbers {an}?n =1 is: 1 x 1!, 2 x 2!, 3 x 3!, The two signs outside the bracket of an represent the range of the sequence. The bottom one is where the sequence begins and the one above is where it should end. Since it is stated the sequence starts from n = 1, therefore the first term, a1 = 1 x 1!, the second term, a2 = 2 x 2! and the third term, a3 = 3 x 3! The ! sign after the numbers is called a factorial notation. The notation basically means the product of all the numbers from 1 to the number with the notation. For example: 3! = 1 x 2 x 3 = 6 5! = 1 x 2 x 3 x 4 x 5 = 120 ? n! = 1 x 2 x 3 x 4 x x n 2! x 3 = 1 x 2 x 3 = 3! = 6 ? n! x (n + 1) = (n + 1)! Going back to the investigation, to find the nth term of the sequence, the steps are shown below: a1 = 1 x 1! = 1 a2 = 2 x 2! = 2 x 1 x 2 = 4 a3 = 3 x 3! = 3 x 1 x 2 x 3 = 18 a4 = 4 x 4! = 4 x 1 x 2 x 3 x 4= 96 . . . ? an = n x n! This is because looking at the sequence, I noticed that there is a similarity in each term. For an, when n = 1, the calculation will be 1 x 1!; when n = 2, the calculation will be 2 x 2!. Therefore from this pattern, the formula to find the nth term is: an = n x n! Let Sn = a1 + a2 + a3 + a4 + + an The term Sn means the summation of all the numbers in the sequence from the first term to the nth term. The mathematical explanation is shown above. For example, a sequence of numbers is 1,2,3,4,5,6,. S1 = 1 S2 = 1 + 2 = 3 S6 = 1 + 2 + 3 + 4 + 5 + 6 = 21 In the sequence that I am investigating, I am told to find Sn for different values of n: S1 = a1 = 1 x 1! = 1 S2 = a1 + a2 = 1 x 1! + 2 x 2! = 1 + 4 = 5 S3 = a1 + a2 + a3 = 1 x 1! + 2 x 2! + 3 x 3! = 1 + 4 + 18 = 23 S4 = a1 + a2 + a3 + a4 = 1 x 1! + 2 x 2! + 3 x 3! + 4 x 4! = 1 + 4 + 18 + 96 = 119 S7 = a1 + a2 + a3 + a4 + a5 + a6 + a7 = 1 x 1! + 2 x 2! + 3 x 3! + 4 x 4! + 5 x 5! + 6 x 6! + 7 x 7! = 1 + 4 + 18 + 96 + 600 + 4320 + 35280 = 40319 By using the information above, I will try to conjecture an expression for Sn. I am going to put all the datas on a table to see if there is any significant discovery. n 1 2 3 4 5 6 an 1 4 18 96 600 4320 Sn 1 5 23 119 719 5039 n! 1 2 6 24 120 720 From the table, I noticed that an is always the difference between n! and (n + 1)! The mathematical expression is: an = (n + 1)! n! Since (n + 1)! seems useful for the investigation, I did another table too: n 1 2 3 4 5 6 (n + 1)! 2 6 24 120 720 5040 Sn 1 5 23 119 719 5039 Looking at the row of (n + 1)! and Sn, there is a constant difference of 1 between them. Therefore I added the row of Sn to the second table (in black). When n = 1, (n + 1)! = 2 ; Sn = 1 n = 2, (n + 1)! = 6 ; Sn = 5 . . . n = 6, (n + 1)! = 5040; Sn = 5039 According to what I have discovered, Sn can be express mathematically in this way: Sn = 1 x 1! + 2 x 2! + 3 x 3! + + n x n! = (n + 1)! 1 To prove that my expression is right, I am using mathematical induction to verify the given result: Pn : 1 x 1! + 2 x 2! + 3 x 3! + + n x n! = (n + 1)! 1 Pk : 1 x 1! + 2 x 2! + 3 x 3! + + k x k! = (k + 1)! 1 If Pk+1 is true the result should be (k + 1 + 1)! 1 = (k + 2)! 1 Pk+1 : 1 x 1! + 2 x 2! + 3 x 3! + + k x k! + (k + 1) x (k + 1)! = (k + 1)! 1 + (k + 1) x (k + 1)! = (k + 1)! [(k + 1) + 1] 1 = (k + 1)! (k + 2) 1 = (k + 2)! 1 ? Pk+1 is true. P1 : LHS = 1 x 1! = 1 RHS = (1 + 1)! 1 = 1 ? P1 is true. ? Pn is true for all positive integers n. Already I have derived the formula an = (n + 1)! n! from the first table. But there is still another way to derive it just from the original formula: an = n x n! = (n + 1 1) x n! = (n + 1) x n! n! = (n + 1)! n! Sn = a1 + a2 + a3 + a4 + a5 + + an = (1 + 1)! 1! + (2 + 1)! 2! + (3 + 1)! 3! + (4 + 1)! 4! + (5 + 1)! 5! + + (n + 1)! n! = 2! 1! + 3! 2! + 4! 3! + 5! 4! + 6! 5! + + (n + 1)! n! At this step, I can see that many numbers cancel out each other except (-1) and (n + 1)! as it goes on to the last term: = -1! + (n + 1)! = (n + 1)! 1 Alternatively, I have used another method to prove of my conjecture for Sn: Sn = 1 x 1! + 2 x 2! + 3 x 3! + + n x n! = (n + 1)! 1 Let cn = an + an+1 According to what I have done before, an = (n + 1)! n! ? cn = (n + 1)! n! + (n + 1 + 1)! (n + 1)! = (n + 1)! n! + (n + 2)! (n + 1)! = (n + 2)! n! To simplify it: cn = (n + 2)! n! = (n + 2) (n + 1) (n!) n! = n! [(n + 2) (n + 1) 1] = n! (n2 + 2n + n + 2 1) = n! (n2 + 3n + 1) Tn = c1 + c2 + c3 + c4 + c5 + + cn To investigate Tn for different values of n, a table is drawn below showing all the values contribute to Tn: n 1 2 3 4 5 6 an : n x n! 1 4 18 96 600 4320 an+1 : (n + 1) x (n + 1)! 4 18 96 600 4320 35280 (n + 1)! 2 6 24 120 720 5040 (n + 2)! 6 24 120 720 5040 40320 n! 1 2 6 24 120 720 cn : (n + 2)! n! 5 22 114 696 4920 39600 T1 = c1 = 1 T2 = c1 + c2 = 5 + 22 = 27 T3 = c1 + c2 + c3 = 5 + 22 + 114 = 141 T4 = c1 + c2 + c3 + c4 = 5 + 22 + 114 + 696 = 837 T5 = c1 + c2 + c3 + c4 + c5 = 5 + 22 + 114 + 696 + 4920 = 5757 T6 = c1 + c2 + c3 + c4 + c5 + c6 = 5 + 22 + 114 + 696 + 4920 + 39600 = 45357 Tn 5 27 141 837 5757 45357 Looking at the column of (n + 1)!, (n + 2)! and Tn, when I add (n + 1)! to (n + 2)!, there is a constant difference of 3 between the sum and Tn. According to what I have found out, Tn can be express mathematically like this: Tn = (1 + 2)! 1! + (2 + 2)! 2! + (3 + 2)! 3! + (n + 2)! n! = (n + 1)! + (n + 2)! 3 Tn = c1 + c2 + c3 + c4 + c5 + + cn-1 + cn ? cn = (n + 2)! n! Tn = (1 + 2)! 1! + (2 + 2)! 2! + (3 + 2)! 3! + (4 + 2)! 4! + (5 + 2)! 5! + + (n 1 + 2)! (n 1)! + (n + 2)! n! = 3! 1! + 4! 2! + 5! 3! + 6! 4! + 7! 5! + + (n + 1)! (n 1)! + (n + 2)! n! At this step, I can see that many numbers cancel out each other except (-1!), (-2!), [(n + 1)!] and [(n + 2)!] as it goes on to the last term: = (n + 1)! + (n + 2)! 1! 2! ? Tn = (1 + 2)! 1! + (2 + 2)! 2! + (3 + 2)! 3! + (n + 2)! n! = (n + 1)! + (n + 2)! 3 In conclusion, throughout the investigation, I have used different methods to find out patterns of the sequences and successfully conjecture expressions for different sequences. Moreover, to prove the conjecture, I used not only by mathematical induciton, but also another method which I carried out for the last part. The 2 main conjectures I have made is: Sn = (n + 1)! 1 Tn = (n + 1)! + (n + 2)! 3 And both of the expressions are true for all positive integers n.