Calculation: 12! + 23! + 34! + +99100! (the final result can be expressed by factorial)

Calculation: 12! + 23! + 34! + +99100! (the final result can be expressed by factorial)

12!+23!+34!+… +99100!=（1-12!）+（12!-13!）+（13!-14!）+… +（199!-1100!）=1-12!+12!-13!+13!-14!+… +199!-1100!=1-1100!
For any natural number n greater than 1, 1 * 2 * 3 * n is defined as n! And read as the factorial of N. calculation: * 1 + 2! * 2 + 3! * 3. * 9
The original formula = 1! (2-1) + 2! (3-1) + 3! (4-1) + 4! (5-1) + +8!(9-1)+9!(10-1)＝2!-1!+3!-2!+4!-3!+5!-4!+…… +9!-8!+10!-9!＝10!-1!=10!-1=3628800-1=3628799
There must be a prime between the factorials of natural number N and N?
The natural number n is required to be n > = 3
It is proved that for any natural number n (n > = 3), there must be prime between N and.
Consider n! - 1
If it is prime, the condition is satisfied
Otherwise
It must contain prime factors except 2 ~ n
There must be, like three
It can be proved by mathematical induction
1 seems to be a natural number, but not a prime number....
Given the function f (x) = loga (1 + x) - loga (1-x) (a > 0 and a ≠ 1), 1 find the domain of definition of function f (x); 2 prove that function f (x) is an odd function
f(x)=loga[(1+x)/(1-x)]
(1+x)/(1-x)>0
(x+1)(x-1)
F (x) is defined as (- 1,1)
f（-x）=loga（1-x）-loga（1+x）=-f（x）
So f (x) is an odd function.
Given the function f (x) = x ^ 2 + 2aX + 3, find the expression f (a) of the minimum value of function f (x) in [- 1,1]
f(x)=x^2+2ax+3
f '(x)=2x+2a=2(x+a);
So when x = - 1, Fmin = 4-2a
-1
f(x)=(x+a)²+3-a²;
When A1, f (x) min = f (- 1) = 4-2a? f(x)=(x+a)²+3-a²; a
Given the function f (x) = loga (x + b) / X-B (a > 1, b > 0) 1, find the domain 2 'to judge the parity
Given function f (x) = loga (x + b) / X-B (a > 1, b > 0)
1 find the domain of definition
2 'judge parity
3-judgement monotonicity
(1) Domain: (x + b) / (X-B) > 0
So: x > b or X
If the function y = 1 / 3x3-1 / 2ax2 + (A-1) x + 1 is a decreasing function in the interval (1,4) and an increasing function in the interval (4, + ∞), find the value of the real number a
y=1/3x³-1/2ax²+(a-1)x+1
y'=x²-ax+a-1
∵ is a decreasing function in the interval (1,4) and an increasing function in the interval (4, + ∞)
When x = 4, y '= 0
That is, 16-4a + A-1 = 0
∴3a=15,
∴a=5
In this case, y '= (x-4) (x-1)
In line with the theme
∴a=5
Function f (x), X ∈ R, if for any real number x1, X2, f (x1 + x2) + F (x1-x2) = 2F (x1). F (x2), we prove that f (x) is even function
Take - X and X as x1, x2
F (x-x) + F (x + x) = 2F (x). F (- x)
-->F(0)+F(2X)=2F(X).F(-X) （1）
Change X1 and X2 again
F (- 2x) + F (- x + x) = 2F (x)
-->F(-2X)+F(0)=2F(X).F(-X) （2）
From (1) (2), f (- 2x) = f (2x)
So f (x) is even function
Let X1 = 0, X2 = 0, we can get 2F (0) = 2F (0) ^ 2, so f (0) = 0 or 1.
Let X1 = 0 again, we can get f (x2) + F (- x2) = 2F (0) f (x2). Substituting f (0) = 0 or 1, we can get f (x2) = 0, or F (x2) + F (- x2) = 2F (x2). The proof is over.
The function f (x) = 1 / 3x3 + 1 / 2ax2 + BX has an extreme point in the interval (- 1,2), (2,3), and the value range of a-4b is obtained
Function f (x), X belongs to R, if for any real number x1, X2 has f (x1 + x2) + F (x1-x2) = 2F (x1) f (x2), prove that f (x) is even function
Let X1 = x2 = 0
Then 2F (0) = 2F (0) & sup2;
If f (0) = 0
Then let x2 = 0
2f(x1)=0
Then f (x) is 0 for any value
Obviously, f (x) is an even function
If f (0) = 1
Let X1 = 0
Then f (x2) + F (- x2) = 2F (x2)
f(-x2)=f(x2)
It is also shown that f (x) is an even function
In conclusion, f (x) is an even function