Series on trot

1x1+1=1
2x1+1=2
3x2x1+1=7
4x3x2x1+1=25 5x4x3x2x1+1=121
6x5x4x3x2x1+1=721

Factorial of natural nos+1

2!+1=3

3!+1=7

4!+1=25

5!+1=121

6!+1=721

It can be solved recursively as well. Through induction it can be proved that $a_n = (a_{n-1})n - (n-1)$ $\forall n \in \mathbf{N}$ such that $n \geq 2$ is equal to $f(n)=n!+1$ :

let $a_1 = 2$

We prove the base case by showing this relation holds true for $n=2$ :

$a_2 = a_1 * 2 - (2-1)$ $\hspace{10mm}$ $f(2)=2! +1$

$a_2 = 3$ $\hspace{40mm}$ $f(2)=3$

Now we assume k is true for all k such that:

$a_k = (a_{k-1})k - (k-1)$ $\hspace{20mm}$ $f(k)=k!+1$

We can rearrange the terms in the $a_k$ recursive formula to give:

$a_k = (a_{k-1} - 1)k + 1$

Since we are trying to prove the equivalence of the two formulas, we can clearly write:

$(a_{k-1} - 1)k + 1$ = $k!+1$

We will try to prove this is true for k+1. We make cancel the 1 term on both sides of the equation and it will be enough to prove the equivalence of the formulas by proving $a_k = k! + 1$ , as we assumed earlier:

$(a_{k} - 1)(k+1)$ = $(k+1)!$

$(a_{k} - 1)(k+1)$ = $(k+1)(k)!$ $\hspace{20mm}$ divide by (k+1)

$a_k - 1 = k!$

$a_k = k! +1$

...

I thought this was a fun exercise in math induction more than anything, but the answer still remains. The 6th term of this recursive formula is $\textbf{721}$ .

2!+1=3 3!+1=7 4!+1=25 5!+1=121 6!+1=721

The series goes as

1!+1

2!+1

3!+1

4!+1

5!+1

and then 6!+1 which is 721

3=2×2-1=2×2-(2-1)
7=3×3-2=3×3-(3-1)
25=7×4-3=7×4-(4-1)
121=25×5-4=25×5-(5-1)
121×6-5=721

(previous number)n-m , n=2,3,4,5,6,...m=1,2,3,4,5,... 2(2)-1=3 3(3)-2=7 7(4)-3=25 25(5)-4=121 121(6)-5=721

Dots are cleverly misleading. :)

My own arithmetic is poor, consequently I resort to Python. This was my last experiment.

>>> for i in range(1,7):

... factorial(i)+1

...

2

3

7

25

121

721

It this serie: (n! +1) n=1 -> (1+1) = 2. n=2 -> (2+1) = 3. n=3 -> (6+1) = 7. n=4 -> (24 +1) =25. n=5 -> (120 +1) = 121. n=6 -> (720 +1) = 721 ... ...

2-3=1=(1*0!^(1/2))^2

7-3=4=(2*1!^(1/2))^2

25-7=18=(3*(2!^(1/2)))^2

121-25=96=(4*(3!^(1/2)))^2

x-121=(5*(4!^(1/2)))^2

x=721

a(5)=(5+1)!+1;a(5)=6!+1;a(5)=720+1;a(5)=721.i take up series starting from 3 and found the 5th term.