1. Recursive formula: a(n+ 1)=a(n)+3b(n), b(n+ 1)=a(n)+b(n). Let a(n)/b(n)=k, then a (n+1)/b (n+1) = (k+1), and the direct expression is: k = (k+3)/. It is proved that if p(n)=a(n)/b(n), then p (n+1) = (p (n)+3)/(p (n)+1), and (P(n)+√3)/ is available or not.

2.q= 1/2, and the molecule is 14d? Let q=a/b, which means 14b? /(a？ +ab+b？ ) is an integer because b? And (a? +ab+b？ ) coprime, so (a? +ab+b？ ) is the divisor of 14, just add them together.

3. Put forward log2(n), the original formula is = (n+2) log2 (1+2/n)-2 (n+1) log2 (1+/n).

=1/ln (2) * ((n+2) * (2/n)-2 (n+1) (1/n)) = 0, and Taylor expansion is the last step. After computer verification, the result is correct.

4. Inductive proof

5.( 1) direct mathematical induction, which is proved to be easy by using the increase and decrease interval of f(x)=x+ 1/x, √ (2n+2)-√ 2n = 2/(√ (2n+2)+√ 2n); The sum of 2+ 1/(2n+C) and series 1/(2n+C) is infinite, and there is no upper bound c at all, which is a direct contradiction.

6.( 1) Expand the original recursive formula into (2a(n+ 1)-7a(n))? =45a(n)？ -36, get a (n+ 1)? -7a(n+ 1)a(n)+a(n)？ +9=0，a(n)=(7a(n+ 1)-√( 45a(n+ 1)？ -36))/2, so a(n- 1)=(7a(n)-√(45a(n)? -36))/2, so a (n+1)+a (n-1) = 7a (n). (2) Direct formula, a (n+1) a (n)-1= (3a (n)+√ (5a (n)? -4))? /4。

7. By directly finding the general formula of series, it can be proved that an+a(n+ 1)+2 is ((3+√ 5)/2) (n-1)+((3-√ 5)/2) (n-1).

8.( 1) It's easy. (2) it is enough to prove a & gtN/(n+ 1) by induction, which is easy to prove, because even1-1(3n+1) can be estimated >: an>1-/.

9. 1-2(x+y)/( 1+x)( 1+y)=( 1-x)( 1-y)/( 1+x)(。

10. Actually, it is1/a1+/a2+...+1/an+1a2 ... An =1,so an =

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