Question

数学大神求解啊

Answer 1

追答

接下来，就靠你自己了，错了请愿谅

Answer 2

设等比数列各项比为x
则 a1 = a2/x = 2/x , a3 = a2*x = 2x,
s3 = a1+a2+a3 = 2/x+2+2x = 7
解等式 x=2.
可知 an = 2^(n-1)

(2) an = 2^(n-1) 则 an+1 = 2^n
带入等式 可得bn = n
数列 {1/(bn+bn+2)} = 1/n*(n+2) = 1/2*(1/n - 1/n+2)
Tn = 1/2(1/1-1/3) + 1/2(1/2-1/4) + 1/2(1/3-1/5) + 1/2(1/4-1/6) .......
= 1/2(1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7......)
= 1/2(1+1/2-1/n+1 - 1/n+2)
= 3/4 - 1/n+1 - 1/n+2
因为n+2 > n+1 > 0
所以1/n+1 > 1/n+2 >0
所以Tn = 3/4 - (大于0的数)
所以Tn < 3/4

Answer 3

高等数学，这几个题目貌似很眼熟，只是多年没拿起，有点生疏了，小伙努力考试吧

Answer 4

追答

如果觉得好，请采纳谢谢