Find the general formula 1, 11111, 1111, 11111

Find the general formula 1, 11111, 1111, 11111

an=1+10+100+…… +10^(n-1)
So it's the sum of isometric tree columns
So an = 1 * (1-10 ^ (n) / (1-10) = (10 ^ n-1) / 9
F(x)=(10^n-1)/9 n=1,2,3,4,5
The general formula is (10 ^ n-1) / 9
Wire 2.5 square column, such as a 2.5 square single core copper wire, use a ruler to measure how thick the copper core should be
Expression of sectional area. Mm mm
2.5 diameter 1.78
4mm Diameter 2.2
16 square meter cable diameter?
The core wire is composed of 7 wires with diameter of 1.7
Given Tan ^ 2A = 2tan ^ 2B + 1, prove: sin ^ B = 2Sin ^ 2a-1
Tan ^ 2A = 2tan ^ 2B + 1sin ^ 2A / cos ^ 2A = 2 * sin ^ 2B / cos ^ 2B + 2-1sin ^ 2A / cos ^ 2A + 1 = 2 * sin ^ 2B / cos ^ 2B + 21 / cos ^ 2A = 2 / cos ^ 2bcos ^ 2B = 2 * cos ^ 2a, so 1-sin ^ 2B = 2 (1-sin ^ 2) = 2-2sin ^ 2a is sorted as sin ^ B = 2Sin ^ 2a-1
Change the tangent to the chord, use the angle doubling formula
We know that M is a real number, and the set M = {x ﹥ x ﹥ 178; - (2m-1) x + (m-2) (M + 1)
1) The inequality in M is reduced to: (x-m-1) (x-m + 2)
Given that a belongs to R, the function f (x) = A / x + inx-1, G (x) = (inx-1) e ^ x + X (where e is the base of the logarithm of natural numbers)
(1) To find out whether the minimum value (2) of function f (x) in the interval (0, e) exists, and whether the real number x0 belongs to (0, e)], so that the tangent of curve y = g (x) at the point x = x0 is perpendicular to the y-axis
(1) F (x) = A / x + inx-1, the domain is (0, + ∞)
F (x) '= - A / X & # 178; + 1 / x = 0, the solution is x = a
① When a ≤ 0
F (x) '= - A / X & # 178; + 1 / x > 0 holds that f (x) is monotonically increasing in the domain of definition and cannot get the minimum value
② 0e hour
The minimum value a / E when x = e
（2）g(x0)‘=0
g(x0)'=e^x0(lnx0+1/x0-1)+1=0
g''(x)=(-1/x^2+2/x+lnx-1)e^x=h(x)e^x
H '(x) is always greater than 0, H (1) = 0
That is g '(x) ≥ G' (1) = 1 > 0
So there is no x0 such that G '(x0) = 0
Derivation
Let the common ratio of the equal ratio sequence {an} be q, the sum of the first n terms be Sn, and A1 > 0. If S2 > 2A3, then the value range of Q is ()
A. (-1，0)∪(0，12)B. (-12，0)∪(0，1)C. (-∞，-1)∪(12，+∞)D. (-∞，-12)∪(1，+∞)
From the meaning of the question, we can get A1 > 0, and a1 + a1q > 2a1q2, that is 2q2-q-1 < 0, that is (2q + 1) (Q-1) < 0. The solution is - 12 < Q < 1, and Q ≠ 0. The value range of Q is (- 12, 0) ∪ (0, 1), so we choose B
Given Tan ^ 2A = 2tan ^ B + 1, prove sin ^ 2B = 2Sin ^ 2a-1
That is sin & sup2; a / cos & sup2; a = 2Sin & sup2; B / cos & sup2; B + 1sin & sup2; ACOS & sup2; b = 2Sin & sup2; bcos & sup2; a + cos & sup2; ACOS & sup2; bsin & sup2; a (1-sin & sup2; B) = 2Sin & sup2; B (1-sin & sup2; a) + (1-sin & sup2; a) (1-sin & sup2; b) Sin & sup2; a
If f (x) = (M & # 178; - m-1) x ^ m & # 178; - 2m-1 is an increasing function in the interval (0, + ∞), then the value set of real number m is:
Process more detailed, I just went to high school, the steps to be detailed. Thank you
If the function is an increasing function, then M is not equal to 0
In addition, if the function is an increasing function at (0, + ∞), then
(m²-m-1)>0
The solution to the inequality is m > (1 + radical 5) / 2 or m
Given the function f (x) = (e ^ X / a) + (A / e ^ x), (a ＞ 0) is an even function on R. (e seems to be the base of the logarithm of natural number de)
(1) The value of A
(2) It is proved that the function f (x) is an increasing function on [0,1]
1) If f (x) is an even function on R, then f (- x) = f (x) leads to a = 1
2) Let f (x) = e ^ x + 1 / e ^ x, let X1 and X2 belong to [0,1] and x1