There is a column of numbers, arranged according to a certain rule: - 1, 2, - 4, 8, - 16, 32, - 64 Is it true that the sum of three adjacent ones is 1224? Please give reasons

There is a column of numbers, arranged according to a certain rule: - 1, 2, - 4, 8, - 16, 32, - 64 Is it true that the sum of three adjacent ones is 1224? Please give reasons

Let the middle number be (- 1) n × 2N-1, then the first number be (- 1) n-1 × 2n-2, (- 1) n + 1 × 2n, ∵ the sum of the three adjacent numbers be 1224, ∵ (- 1) n × 2N-1 + (- 1) n-1 × 2n-2 + (- 1) n + 1 × 2n = 1224, ∵ (- 1) n

The number of regularly arranged columns: 2, 4, 6, 8, 10, 12. Each term can be expressed by formula 2n (n is a positive integer). The number of regularly arranged columns: 1, - 2, 3, - 4, 5, - 6, 7, - 8. What formula do you think can be used to express each term?

n（-1)^(n+1)

If A1 = 1, then a2007 = () (2007 Fujian civil servants real topic) the analysis is like this: from an + 1 = 1-1 / (an + 1), we can get: 1 / (an + 1) 1 / an, that is {1 / an} is an arithmetic column with tolerance of 1, the first item is 1 / A1 = 1, then 1 / a2007 = 1 / 2007 I don't understand, Why can 1 / (an + 1) 1 / an be deduced from an + 1 = 1-1 / (an + 1), and why {1 / an} is an arithmetic sequence with tolerance of 1?

An + 1 = 1-1 / (an + 1) = (an + 1) / (an + 1) - 1 / (an + 1) = an / (an + 1), so 1 / (an + 1) = (an + 1) / an = 1 + 1 / an, then 1 / (an + 1) - 1 / an = 1

There is a column of numbers a1a2a3... An starting from the second number, each number is equal to the quotient of the number before it minus 1 divided by the number before it, if A1 = 2
So what is A2008

A2 is one in two
A3 equals - 1
A4 equals two
A5 is one in two cycles
Dividing 2008 by 3 is the remainder, so A2008 is 1 / 2