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∴ 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.
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:
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
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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:
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
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.
