For any natural number n greater than 1, it is proved that: (1 + 1 / 3) (1 + 1 / 5) (1 + 1 / 7), (1 + 1 / 2n-1) > radical 2N-1 / 2

For any natural number n greater than 1, it is proved that: (1 + 1 / 3) (1 + 1 / 5) (1 + 1 / 7), (1 + 1 / 2n-1) > radical 2N-1 / 2

With mathematical induction, n = 2, hold
Suppose n = k, the proposition holds: (1 + 1 / 3) (1 + 1 / 5) (1 + 1 / (2k-1)) > radical (2k + 1) / 2
All you need is a certificate
(1 + 1 / 2K + 1) (radical (2k + 1) / 2) >
The root sign (2k + 3) / 2 is enough
Namely (2k + 3) / (2k + 1) > radical (2k + 3) / radical (2k + 1)
Because the number greater than 1 is smaller than the original one,
So (2k + 3) / (2k + 1) > radical (2k + 3) / radical (2k + 1) holds, and then the original question is proved
Let Tan α = - 1 / 2, calculate 1 / sin α - sin α cos α - 2cos & sup2; α
Use a instead
It should be sin & sup2; a
sina/cosa=tana=-1/2
cosa=-2sina
Substituting Sin & sup2; a + cos & sup2; a = 1
sin²=1/5
cos²a=4/5
sinacosa=sina(-2sina)=-2sin²a=-2/5
Original formula = 1 / (1 / 5 + 2 / 5-8 / 5) = 1
If y = f (x) is a decreasing function on R and the image of y = f (x) passes through points a (0,1) and B (3, - 1), then the solution set of the inequality | f (x + 1) | < 1 is______ .
Y = f (x) is a decreasing function on R, and the image of y = f (x) passes through points a (0,1) and B (3, - 1), so the solution set of | f (x) | 1 is If {x | 0 ＜ x ＜ 3}, the image of the function y = | f (x + 1) | corresponding to the inequality | f (x + 1) | 1 can be regarded as the image of y = | f (x) | which is shifted one unit to the left, then the solution set of the inequality | f (x + 1) | 1 is: {x | - 1 ＜ x ＜ 2}, so the answer is: (- 1, 2)
Prove: (Sin & sup2; α + Tan α * Tan α / 2 + cos & sup2; α) * sin α & sup2. / 2cos α = Tan α
[Sin & # - 178; α + Tan α * Tan (α / 2) + cos & # - 178; α] * Sin & # - 178; (α / 2) * cos α = [1 + Tan α * (1-cos α) / sin α] * (1-cos α) / 2 * cos α = [1 + (1-cos α) / cos α] * (1-cos α) / 2 * cos α = (1-cos α) / 2 you check the question and see what's wrong
More than ten years, I can't remember the transformation of trigonometric formula. This should be very simple. Just use trigonometric formula transformation. You can have a try
If the solution of the two-point inequality (| 2, f) on the graph is known to be a minus function (| 2)
A. （-1，2）B. （-∞，1）∪（4，+∞）C. （-∞，-1）∪（2，+∞）D. （-∞，-3）∪（0，+∞）
∵ f (X-2) | > 2, ∵ f (X-2) > 2 or F (X-2) ＜ - 2, and ∵ a (0, - 2), B (- 3, 2) are two points on the image, ∵ f (0) = - 2, f (- 3) = 2, ∵ function f (x) is a decreasing function on R,