Find a formula for this series Pleeeeez! 6, -7, 8, -9, 10...?

Answer 1

My suggestion:

Given
n = 1, x = 6
n = 2, x = -7
n = 3, x = 8
n = 4, x = -9 and so on.

x = (-1)^(n+1) . (n+5) where n is Integer >= 1

Answer 2

You can do it with discrete mathematics (A type of math that only uses whole numbers, rather than fractions):

X = (5+Y) * -((Y-1)%2)

Where:

* is multiply
% is modulus

In discrete mathematics, modulus means "The remainder after a division", so:

1%2=1 (Since if you divide 1 by 2, you get 0 remainder 1)
2%2=0 (2 / 2 = 1 remainder 0)
3%2=1 (2 / 3 = 1 remainder 1)
etc...

Edit:

KZslot's answer is probably better, since "-((Y-1)%2)" is equal to "-1^(Y-1)", where ^ means "to the power of"

Answer 3

subtract (2x+1) from the previous number: replace the variable with the previous number and do the math within the brackets. Then subtract that answer from the previous number.

ex. (2(6)+1)= (12+1)= 13, 6-13= -7
I think this works throughout the sequence.

Answer 4

You could just say that the numbers increase by 1 each time (6,7,8,9,10,...) but the signs alternate (+,-,+,-,+...).

You could also look at the problem as 2 series mixed togethet. The odd-numbered terms are 6,8,10,... Those clearly increase by 2 each time. The even-numbered terms are -7, -9,... Those decrease by 2 each time. You can continue the patterns as far as you need to.

Answer 5

-8, +15, -7, +19

Answer 6

1)By magnitude, the numbers are consecutive, starting at "6":

"n + 6" {from n = 0 to infinity}



2)The odd numbers are negative, the evens are positive
"(-1)^(n)" will give "-1" for n being odd, and "1" for n being even

Put it together:
Series = (-1)^(n) * (n) {from n = 0 to infinity}

Answer 7

It looks like a pattern to me... would the next few ones be -11, 12, -13, 14...

It's One minus the previous answer :-) I think...

Answer 8

- 11

Answer 9

-11, 12, -13 it's a pos nr then a neg in cronological order

Answer 10

(-1)^n * (n+6)

(assuming you count from 0)