Consider the sequence

1, 2, 4, 8, ...

and prove that any term in the sequence is 1 greater than the sum of the
previous terms.

Okay

n = 1 and the first term is n .

The second term is 2n but set the second term equal to the sum of the
previous terms as

2n - 1 = n .

Now going to the next term the left side of this equation will have an
n-value multiplied by 2 while the right side of this equation will have
the addition of an n-value that is 2 times the previous n-value of the
right side.

But note that the n-value on the left side of the equation is 2 times
greater than the n-value on the right side of the equation.

So going to the next step in the sequence as

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

the right side of the equation is increased by a value equal to the
n-value of the previous step of the left side of the equation and the left
side of the equation has the n-value of the previous step multiplied by 2
or in other words increased in value by the amount of the previous
n-value. Both sides of the equation will always increase
by the same amount with each step of multiplying on the left side and with
each step of addition on the right side.

With the premise true at the second term of the sequence it will also be
true at any of the larger terms of the sequence.

And that's all copied from a previous KBH post...

-------------------------------------------------------

Now consider the sequence

1, 4, 16, 64, ...

Now for a term to relate to the sum of the previous terms

1 needs to relate to 4
5 needs to relate to 16
21 needs to relate to 64

And the second term 4 is 3 greater than the sum of its previous terms, the
third term 16 is 11 greater than the sum of its previous terms, and the
fourth term 64 is 43 greater than the sum of its previous terms.


So look at the pattern of

3, 11, 43

and note that 11 is one less than a multiple of 3 and that 43 is 1 less
than a multiple of 11. But doesn't really help...except that the pattern
began with 3 .

Now relate 1 to 4 as

(1*3) + 1 = 4

And relate 5 to 16 as

(5*3) + 1 = 16

Finally relate 21 to 64 as

(21*3) + 1 = 64 .

In other words any term in the sequence can be a multiple of 3 plus 1 .

Now remember that the values 1, 5, and 21 were the sums of the pervious
terms. So make those values the unknowns to solve for the sums of the
previous terms.

(x*3) + 1 = 4
x = 1

(x*3) + 1 = 16
x = 5

and

(x*3) + 1 = 64
x = 21

So the formula for the 1, 4, 16, 64, ... sequence is

x = (n-1) / 3

where n is the upper and non-inclusive term bound of the sum of terms from
the first term.


Now since the 1, 2, 4, 8, ... sequence has any term as 1 greater than the
sum of its previous terms then the formula for that sequence is

x = (n - 1) / 1 .


Then if I go to conventional terminology these
sequences are

a, ar, ar^2, ar^3, ...

as

1, (1*2), (1*2)^2, (1*2)^3, ...

and

1, (1*4), (1*4)^2, (1*4)^3, ...


Now I'm using n to represent a*r^power and I'm using x to represent a sum
of terms.

So the general formula for the sum of the previous terms of a geometric
progression that begins with 1 and where a = 1 and where all terms are
positive integers is

x = (n - 1) / (r - 1) .


And that's just how it was done...and represents an additional hour of
work on the subject.

Consider the sequence

1, 2, 4, 8, ...

and prove that any term in the sequence is 1 greater than the sum of the
previous terms.

Okay

n = 1 and the first term is n .

The second term is 2n but set the second term equal to the sum of the
previous terms as

2n - 1 = n .

Now going to the next term the left side of this equation will have an
n-value multiplied by 2 while the right side of this equation will have
the addition of an n-value that is 2 times the previous n-value of the
right side.

But note that the n-value on the left side of the equation is 2 times
greater than the n-value on the right side of the equation.

So going to the next step in the sequence as

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

the right side of the equation is increased by a value equal to the
n-value of the previous step of the left side of the equation and the
left side of the equation has the n-value of the previous step multiplied
by 2 or in other words increased in value by the amount of the previous
n-value. Both sides of the equation will always increase
by the same amount with each step of multiplying on the left side and
with each step of addition on the right side.

With the premise true at the second term of the sequence it will also be
true at any of the larger terms of the sequence.

And that's all copied from a previous KBH post...

-------------------------------------------------------

Now consider the sequence

1, 4, 16, 64, ...

Now for a term to relate to the sum of the previous terms

1 needs to relate to 4
5 needs to relate to 16
21 needs to relate to 64

And the second term 4 is 3 greater than the sum of its previous terms,
the third term 16 is 11 greater than the sum of its previous terms, and
the fourth term 64 is 43 greater than the sum of its previous terms.


So look at the pattern of

3, 11, 43

and note that 11 is one less than a multiple of 3 and that 43 is 1 less
than a multiple of 11. But doesn't really help...except that the pattern
began with 3 .


Oh, more from the notes...

And the second term 4 is 3 greater than the previous term, the third term
16 is 12 greater than its previous term, and the fourth term 64 is 48
greater than its previous term.

So look also at the pattern

3, 12, 48

and the differences between terms are all multiples of 3 .

Now relate 1 to 4 as

(1*3) + 1 = 4

And relate 5 to 16 as

(5*3) + 1 = 16

Finally relate 21 to 64 as

(21*3) + 1 = 64 .

In other words any term in the sequence can be a multiple of 3 plus 1 .

Now remember that the values 1, 5, and 21 were the sums of the pervious
terms. So make those values the unknowns to solve for the sums of the
previous terms.

(x*3) + 1 = 4
x = 1

(x*3) + 1 = 16
x = 5

and

(x*3) + 1 = 64
x = 21

So the formula for the 1, 4, 16, 64, ... sequence is

x = (n-1) / 3

where n is the upper and non-inclusive term bound of the sum of terms
from the first term.


Now since the 1, 2, 4, 8, ... sequence has any term as 1 greater than the
sum of its previous terms then the formula for that sequence is

x = (n - 1) / 1 .


Then if I go to conventional terminology these
sequences are

a, ar, ar^2, ar^3, ...

as

1, (1*2), (1*2)^2, (1*2)^3, ...

and

1, (1*4), (1*4)^2, (1*4)^3, ...

Now I'm using n to represent a*r^power and I'm using x to represent a sum
of terms.

So the general formula for the sum of the previous terms of a geometric
progression that begins with 1 and where a = 1 and where all terms are
positive integers is

x = (n - 1) / (r - 1) .


And that's just how it was done...and represents an additional hour of
work on the subject.