1-1 / 2-1 / [the 2nd power of 2] - 1 / [the 3rd power of 2] -. - what is the number of - 1 / [the nth power of 2]

1-1 / 2-1 / [the 2nd power of 2] - 1 / [the 3rd power of 2] -. - what is the number of - 1 / [the nth power of 2]

1-1 / 2-1 / [2nd power of 2] - 1 / [3rd power of 2] -. - 1 / [nth power of 2]
It can be regarded as the sum of the equal ratio sequence with the first term of 1 and the common ratio of - 1 / 2
1-1 / 2-1 / [2nd power of 2] - 1 / [3rd power of 2] -. - 1 / [nth power of 2]
=1*(1-(-1/2)^n)/(1-(-1/2))
=2/3(1-(-1/2)^n)
Function y = LNX (0
Base E > 1
So LNX is an increasing function
So LNX ≤ ln1 = 0
Range (- ∞, 0]
y=lnx
So the function increases monotonically at (0, positive infinity)
And (0,1) is less than or equal to 0
So the range of y = LNX on (0,1] is (negative infinity, 0]
The second power of two minus two minus the third power of two minus the fourth power of two minus the fifth power of two Minus two to the power of 2011 plus two to the power of 2012
The second power of two minus two minus the third power of two minus the fourth power of two minus the fifth power of two Minus two to the power of 2011 plus two to the power of 2012
=2012 power of 6-2 + 2012 power of 2
=6
Given the function f (x) = ax + LNX, X ∈ (1, e), and f (x) has extreme value, how to find the solution of the range of function f (x)
F '(x) = a + 1 / x = 0 has a solution in (1E)
One
First of all, f '(x) = (AX + 1) / X
Because there is an extremum, Let f '(x) = 0, x = - 1 / A, because x > 0, and because the extremum is on (1, e), so a
Evaluation: X (x + 2Y) - (x + 1) 2 + 2x, where x = 125, y = − 25
X (x + 2Y) - (x + 1) 2 + 2x = x2 + 2XY - (x2 + 2x + 1) + 2x = x2 + 2xy-x2-2x-1 + 2x = 2xy-1. When x = 125, y = − 25, the original formula = 2xy-1, = 2 × 125 × (- 25) - 1, = - 3
Function y = f (x) when x belongs to [a, b], the range is [Ka, KB] (k > 0), then y = f (x) is called k-times function. If f (x) = LNX + X is k-times function
Finding the value range of real number k
F (x) = LNX + X, the domain of definition is x > 0
F (x) is a monotone increasing function in the domain of definition
So f (a) = Ka, f (b) = KB
That is: LNA + a = Ka
lnb+b=kb
That is, a and B are two different roots of the equation LNX + x = KX
k=1+(lnx)/x=g(x)
G '(x) = (1-lnx) / x ^ 2 = 0, the maximum point x = e is obtained
The maximum of G (x) is g (E) = 1 + 1 / E
g(0+)=-∞,g(+∞)=1
So when 1
Simplify the evaluation of the square of 2x-3 / 2Y - (- x + 3 / 1y) x = - 1 / 3, y = - 1 and 2 / 3
Simplify the evaluation of the square of 2x-3 / 2Y - (- x + 1y / 3)
=(2x-3 / 2y-x + 3 / 1y) (2x-3 / 2Y + x-3 / 1y)
=(1y of x-3) (3x-y) x = - 1 of 3, y = - 1 and 2 of 3
=(-1/3+5/9)(-1+5/3)
=2/9*2/3
=4/27
The function y = x ^ 2 + LNX, X belongs to the range of [2, e ^ 2]
X is a positive number
So y = x ^ 2 is an increasing function
LNX is also an increasing function
So y = x ^ 2 + LNX is an increasing function
So x = 2, y min = 4 + LN2
X = e ^ 2, ymax = e ^ 4 + lne ^ 2 = e ^ 4 + 2
X = e ^ 2 is not available
So the range [4 + LN2, e ^ 4 + 2]
First simplify, then evaluate. X (x + 2Y) - (x + 2) ^ 2 + 2x, where x = 2, y = - 1 / 2
X (x + 2Y) - (x + 2) ^ 2 + 2x = x & # 178; + 2XY - (X & # 178; + 4x + 4) + 2x = x & # 178; + 2xy-x & # 178; - 4x-4 + 2x = 2xy-2x-4 = 2x (Y-1) - 4 ∵ x = 2, y = - 1 / 2 ∵ original formula = 2 × 2 × (- 1 / 2-1) - 4 = 2 × 2 × (- 3 / 2) - 4 = - 6-4 = - 10
Find the range of function y = x ^ 2-6x + 8 (1 ≤ x ≤ 6)
Solution y = x ^ 2-6x + 8
=（x-3）^2-1
It belongs to [1,6]
We know that when x = 3, y has a minimum value of - 1
When x = 6, y has a maximum value of 8
So the range of the function is [- 1,8]
The intersection of function and X axis is (4,0) (2,0)
The abscissa of the vertex is - 2A, B = 3
a> 0 opening up, so
When x = 1, y = 3
When x equals 3, y = - 1
When x = 6, y = 8
So y is greater than or equal to - 1 and less than or equal to 8