# Please write a conjecture about each pattern and write the next two patterns: Number One: 1/2, 1/4, 1/8. Number Two: -3, 6, -9, 12.

hala718 | Certified Educator

1) 1/2  1/4  1/8 ....

Let us rewrite:

1/2 , (1/2)^2   (1/2)^3.....

We conclude that the above is a Geometric progression where r = 1/2   and frist term a1= 1/2

==> a1= 1/2

==> a2= 1/2 *1/2

==> a3= 1/2 *(1/2)^2 = (1/2)^3= 1/8

==> a4= 1/2 * (1/2)^3 = (1/2)^4 = 1/16

==> a5 = (1/2)*(1/2)^4 = (1/2)^5 = 1/32

.....

==> an = (1/2)*(1/2)^(n-1) = (1/2)^n  = 1/2^n

==> 1/2, 1/4, 1/8, 1/16, 1/32, ...., 1/2^n

-3, 6, -9, 12 , ....

Also we notice that :

an = (-1)^n *3n

==> a1= (-1)^1 * 3*1 = -3

==> a2= (-1)^2 * 3*2 = 6

==> a3= (-1)^3 *3*3 = -9

==> a4= (-1)^4 * 3*4 = 12

==> a5= (-1)^5 * 3*5 = -15

==> a6 = (-1)^6 *3*6 = 18

.....

-3, 6, -9, 12, -15, 18, ....., (-1)^n * 3n

krishna-agrawala | Student

(1)

The given series of numbers is:

1/2, 1/4, 1/8, ...

This can also be represented as:

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

Thus the nth term of the series can be represented as (1/2^n).

Using this formula for the nth term, the first 5 terms of the series become:

1/2, 1/4, 1/8,1/16, 1/32, ...

(2)

The given series of numbers is:

-3, 6, -9, 12, ...

This can also be represented as:

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

Thus the nth term of the series can be represented as (-1^n)(3n).

Using this formula for the nth term, the first 6 terms of the series become:

-3, 6, -9, 12, -15, 18, ...