作业帮求大神解答~~高中数学等比数列、等差数列的问题,急求!
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求大神解答~~高中数学等比数列、等差数列的问题,急求!
求大神解答~~高中数学等比数列、等差数列的问题,急求!
求大神解答~~高中数学等比数列、等差数列的问题,急求!
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).