Question

求大神解答~~高中数学等比数列、等差数列的问题，急求！

Answer 1

1. Setting n=5 in a_5*a_{2n-5}=2^{2n} yields a_5=2^5. Putting it back yields a_{2m-1}=2^{2m-1} for any m>=1. So the desired number is 1+3+5+...+(2n-1)=n^2.

2. Suppose a_5=x. Then a_4=x/q, a_6=xq, and a_7=xq^2. So 2(1+q^2)=1/q+q. Solving it we get q=1/2 or q=i or q=-i.

3. Computing the first terms of b_n, we see that it is not a geometric progression. The second problem can be solved by the characteristic root method. But I believe the question was wrongly printed.

4. b_1=1 and b_{n+1}*(n+1)=(n+1)*b_n+n+1/2^n. Again, it is solvable but I believe the question was wrongly printed. The corrected version could be a_{n+1}=(1+1/n)a_n+(n+1)/2^n. In other words, it seems that the last two summands should be replaced by (n+1)/(2^n). Here is a hint for solving such questions: for Q4(1), write the recurrence as b_{n+1}-b_n=***, then solve the arithmetic progression b_n; for Q4(2), use the answer of Q4(1).