It is known that f (x) is defined on the set of natural numbers, and for any x ∈ n *, there is f (x) = f (x-1) + F (x + 1), where f (1) = 2008. Ask if f (x) is a periodic function? If f (x) is a periodic function, find one of its periods, and find f (2008)

It is known that f (x) is defined on the set of natural numbers, and for any x ∈ n *, there is f (x) = f (x-1) + F (x + 1), where f (1) = 2008. Ask if f (x) is a periodic function? If f (x) is a periodic function, find one of its periods, and find f (2008)

That is, f (x + 1) = f (x) - f (x-1) Let f (0) = a, ∵ f (1) = 2008 ∵ f (2) = 2008-A, f (3) = - A, f (4) = - 2008, f (5) = A-2008, f (6) = a, f (7) = 2008 ∵ cycle is 62008 = 6 * 334 + 4 ∵ f (2008) = f (4) = - 2008

20n is 1 × 2 × 3 × 4 × The natural number n is the largest______ ．

20n = (2 × 2 × 5) n = 2n × 2n × 5N, where 1 × 2 × 3 × 4 × In 2011 × 2012, the number of 2 is far more than 5, so 1 × 2 × 3 × 4 × How many 5 can be decomposed at most in × 2011 × 2012, that is, the maximum value of N, 2012 ／ 5) + (2012 ／ 25) + (2012 ／ 12

If a, B are natural numbers, and a > b, 2011 = a (A-1) + B, then a =, B=

First, we get that B is even from 2011 = a (A-1) + B, because whether a is odd or even, A-1 is even or odd, a (A-1) is odd, and 2011 is odd, so B must be even. Secondly, 2011 = a (A-1) + B = = = > 2011-b = a (A-1) that is, the product of two continuous natural numbers equals 2011-b, assuming B = 0