Investigating a Sequence of Numbers

S1=1, S2 = 4, S3 = 6, S4 = 24 ...(X)

So I tried to make Sn=n! into Sn=(n+1)! And I realized that an expression for Sn I want to find equals subtracting 1 from (n+1)!.

S1 = 2, S2 = 6, S3 = 24, S4 = 120...

So I could conjecture an expression for Sn

@ My conjecture is Sn = (n+1)! - 1

Part 4

Prove your conjecture by the use of mathematical induction.

Prove that Sn = a1 + ... + an = (n+1)!-1

i ) when n = 1

L.S : S1= a1 = 1 x 1! = 1

R.S : S1 = (1+1)! - 1 = 2! - 1 = 1

Therefore the formula is true for n=1

ii ) Assume the formula is true for n=k

Sk =(k+1)! - 1

iii ) when n = k+1

Sk+1 = (k+2)! - 1

Sk+1=a1+a2+a3+...+ak+ak+1

= (a1+a2+a3+...+ak)+ak+1

= Sk + ak+1 (since Sk= a1+a2+a3+...+ ak)

= (k+1)! - 1+ ak+1

= (k+1)! -1+ (k+1) x (k+1) (As I mentioned in part 1, I could get this formula an = n x n!)

= (k+1)!(k+2) - 1

= (k+2)! - 1 (since(k+2)(k+1)! = (k+2)! If n= k+1 then ak+1=(k+1) x (k+1)!)

If the formula is true for n = k it is also true for n = k+1

But it is true for n = 1

n = 2

n = 3

And so on for all positive values of n.

@ Therefore I have proved that Sn = a1 + ... + an = (n+1)! - 1

Part 5

Show that an can be expressed as (n+1)! - n! and devise an alternative (direct) proof of your conjecture for Sn.

(n+1)!-n! can be expressed as (n+1) x n! - n! by factorial notation(e.g. (3+1)!-3! = 4 x 3! - 3!)

so (n+1)!-n!= (n+1) x n! - n!

= n! x n + n! - n!

= n x n!

As I mentioned in part 1 , an = n x n!

@ Therefore an can be expressed as (n+1)!-n!

But also I could prove that Sn = n x n! = (n+1)! - n! which is my conjecture by using direct proof.

Sn = a1 + a2 + a3 + ... + an

= (1 x 1!) + (2 x 2!) + (3 x 3!) + ... + (n x n!)

= (2! - 1!) + (3! - 2!) + (4! - 3!) +... +{(n+1)!-n!}

= {2!+3!+4!+...+n!+(n+1)!} - (1!+2!+3!+...+n!) (we can cancel)

= (n+1)! - 1!

= (n+1)! - 1

@ By using direct proof, an can be expressed as (n+1)! - n!

Part 6

Let cn = an + an+1 Write an expression for cn in factorial notation and simplify it.

As we know from part 5, an can be expressed as (n+1)!-n! and an+1 can be expressed as (n+2)!-(n+1)!

Therefore cn = an+an+1

= (n+1)! - n! +(n+2)! - (n+1)! (we can cancel)

= (n+2)! - n!

@ Simplified expression of cn = (n+2)! - n!

Part 7

Let Tn = c1 + c2 + ... + cn Investigate Tn for different values of n.

n

cn=(n+2)! -n!

Tn

5

c1=5

2

22

c1+c2=27

3

14

c1+c2+c3=141

4

696

c1+c2+c3+c4=837

5

4920

c1+c2+c3+c4+c5=5757

When n = 1 T1 = c1 = 5

n = 2 T2 = c1+c2 = 22

n = 3 T3 = c1+c2+c3 = 141

n = 4 T4 = c1+c2+C3+C4 =837

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Part 8

Conjecture an expression for Tn

It was very difficult to conjecture an expression for Tn with just values of n.

It was not simple.

So I tried to conjecture by using result from part 6.

As we know from part 6, cn= (n+2)! - n!

Tn = c1 + c2 + c3 +... + cn

Therefore Tn = {(1+2)!-1!} + {(2+2)!-2!} +{(3+2)!-3!}+...+{(n+2)!-n!}

= (3!-1!) + (4!-2!) +(5!-3!)+... + {(n+2)!-n!}

= {3!+4!+5!+... n!+(n+1)!+(n+2)!} - (1!+2!+3!+... n!) (we can cancel)

= (n+1)! + (n+2)! - 1 - 2!

= (n+1)!(n+2)(n+1)! -1 -2!

= (n+1)!(n+1+2) -1 -2!

= (n+1)!(n+3) - 3

@ Therefore my conjecture is Tn = (n+1)!(n+3) - 3

Part 9

Prove your conjecture by any method.

Actually I already proved my conjecture in part 8 by using direct proof.

But also I could prove my conjecture by using mathematical induction.

Prove that Tn = c1+c2+c3+...+cn = (n+1)!(n+3) - 3

i ) when n =1

L.S : T1 = c1 = 3! - 1! = 5

R.S : T1 = (1+1)!(1+3) - 3 = 5

Therefore the formula is true for n = 1

ii ) assume the formula is true for n = k

TK = (k+1)!(k+3) - 3

iii ) when n = k+1

TK+1 = (K+2)!(K+4) - 3

TK+1 = c1+c2+c3+c4+... + ck+ck+1

= Tk + ck +1 (since Tk= c1+c2+c3+c4...+ck+ck+1)

= (K+1)!(K+3)-3 + (k+3)!-(k+1)! (since Tk = (k+1)!(k+3) -3

ck+1 = (k+3)!-(k+1)! By part 6)

= (k+1)!(k+3-1) + (k+3)! - 3

= (k+1)!(k+2) + (k+3)! - 3

= (k+2)!+(k+3)!- 3 (since (k+1)!(k+2) = (k+2)!)

= (k+2)!+ (k+3)(k+2)! - 3 (since (k+3)! = (k+3)(k+2) )

= (k+2)!(k+1+3) - 3

= (k+2)!(k+4) - 3

if the formula is true for n = k it is also true for n = k+1.

But it is true for n = 1

n = 2

n = 3

And so on for all positive values of n.

@ Therefore I have proved that Tn = c1+...+cn = (n+1)!(n+3) - 3