Question

高二上数学等差数列，等比数列题目求教大神

已知数列{an}中，an=n^2+kn(n∈N*)，若数列{an}是递增数列，求实数k的取值范围.（标答是k≥-2）我算出来是≥-1，究竟算出来是哪个？
在△ABC中，若三边成等比数列，最小边为a，求三角形周长L的取值范围。网上有说q＞1是怎么来的呢？
等比数列{an}的前n项和Sn=2^n-1,则a1^2+a2^2+a3^3+…+an^2=_________怎么做？
已知数列｛an｝的前n项和Sn=5^n+t，（t是实数），下列结论正确的是（B）如何判断？
A t为任意实数，｛an}均是等比数列 B当且仅当t=-1时，｛an}是等比数列C当且仅当t=0时,{an}是等比数列 D当且仅当t=-5时，｛an｝是等比数列
等比数列{an}的前n项和为Sn,首项为a1,公比为q(q≠1）,则数列{1/an}的前n项和为______ 求解题方法
已知等比数列{an}中，a2=1,则其3项的和S3的取值范围是_________求解
急急急

Answer 1

1. 将数列看出一个开口向上的二次函数（只取其中的正整数）只要对称轴小于等于1就满足条件，解出k≥-2.

2. 三条边为a、qa、q²a，边长肯定为正，所以q>0,a为最小边，那么qa>a，得出q>1

3. S（n-1）=2^(n-1)-1,an=Sn-S（n-1）=2^(n-1),设bn=an²=2^n，则bn为以2为首项，2为公比的等比数列，前n项和Tn=b1+b2+……+bn=a1^2+a2^2+a3^3+…+an^2=2^(n+1)-2

4. S（n-1）=5^(n-1)+t,an=Sn-S（n-1）=4/5*5^(n-1)  (n≥2)，a1=S1=5+t。当n=1.即4/5*5^(1-1)=5+t，t=-1时，｛an}是等比数列。答案是B

5. 数列{1/an}的首项为1/a1,公比为1/q,则数列{1/an}的前n项和为Sn=1/a1*(1-q^-n)/(1-1/q)

6. 设公比为q（q≠0），则a1=1/q，a2=1，a3=q，由基本不等式得出1/q+q大于等于2，或者小于等于-2，则S3≥3或者S3≤-1

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第3题你的答案不对，标答是1/3(4^n -1)！

追答

不好意思，回答错了，bn=an^2=4^（n-1），

Answer 2

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).

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为啥用英文？