Second, the recursive formula of cumulative commercial law is: an+ 1=f(n)an(f(n) should be integrable). Let n = 1, 2, ... N- 1 to get A2/a1= f. Integrable ∴ an = a 1 f (1) f (2) ... f (n-1) Of course, we need to verify whether a1is applicable to the above formula 2 when n= 1. In the sequence {an}, A 1. N- 1 can get a2/a1= f (1) a3/a2 = f (2) a4/a3 = f (3) ... an/an-1= f (n-/kloc-0
3. The construction method is 1, and the recurrence relation is an+ 1=pan+q (p, q is constant). Thinking: Let the recurrence formula be an+ 1+x=p(an+x) and get an+ 1 = pan+(p-). The solution is x=q/(p- 1), so the recurrence formula can be changed to an+ 1+x=p(an+x) construction sequence {bn}, bn = an+q/(p-1) bn+1. 1(n？ N) If an=2an- 1+3, the recursive formula for finding an can be changed to an+x=2(an- 1+x), an=2an- 1+x, and x=3, so the recursive formula can be changed to an+3 = 2 (an {bn} Bn = an+3bn = 4× 3n-65438q is a constant) Idea: divide an+ 1=pan+qn by qn+ 1 to get an+1= p/qan/qn. Bn=an/qn can get bn+ 1=p/qbn+ 1/q, so we can get bn=f(n) in the sequence {an}, A 1+5/6, AN+65438 by using the above solution. Find an+1= (1/3) an+(1/2) n and divide it by (1/2)n+ 1 to get 2n+1an+/kloc. Bn=2nan can get bn+ 1=(2/3)bn+ 1, so BN = 3-2× (2/3) n2nan = 3-2× (2/3) nan = 3× (1/2) can. Q is constant): let an+2=pan+ 1+qan be transformed into an+2-xan+1= y (an+1-xan), that is, an+2 = (x+y) an+1. Example 5: In the known sequence {an}, A 1 = 1, A2 = 2, An+2 = (2/3). Let an+2 = (2/3) an+1+(1/3) an be transformed into an+2-xan+1= y (an+1-xan), that is, an+2 = (x+) Y=-1/3 to construct the sequence {{bn}}, Bn=an+ 1-an Therefore, the sequence {bn} is a geometric series with the public ratio of-1/3, that is, bn = b1(-/kloc-0. N-1an+1-an = (-1/3) n-1,so we can get an =1+3/4× [1-(- N*)
4. Find an 1 by using the relationship between sn and N and an, and find an by using the relationship between sn and N: When n= 1, when n≥2, an=sn-Sn-1Example 6, the first few of known series and s=n2+ 1 Find {when An = sn-sn-1= n+1-[(n-65438an = 2n-12), and find an idea by using the relationship between sn and an: by using an=sn-sn- 1, we can get the recursive relationship. In the sequence {An}, it is known that sn=3+2an, and finding an is an = sn-sn-1= 3+2an-(3+2an-1) an = 2an-1∴ {an} is a common ratio. Then it is proved by mathematical induction that in the known sequence {an} in Example 8 (National College Entrance Examination in 2002), an+ 1 = A2N-Nan+ 1 = 2, and A 1 = 2, so that an can be found from the known sequence {an}. It is proved by mathematical induction that when n= 1, left = 2, right = 2 and left = right, that is, when n= 1, the proposition holds. That is, if ak=k+ 1, then AK+1= a2k-kak+1= (k+1) 2-k (k+1)+1= k2+.