Given the sequence {a (n)}, a (1) = 1, a (2) = 2, a (n) = a (n-1) - A (n-2), (n ∈ n + and N ≥ 3), then s (100)=

Given the sequence {a (n)}, a (1) = 1, a (2) = 2, a (n) = a (n-1) - A (n-2), (n ∈ n + and N ≥ 3), then s (100)=

a1=1,a2=2,a3=1,a4=-1,a5=-2,a6=-1,
a7=1,a8=2...
This is a sequence with period 6
100=16*6+4
S100=16(a1+a2+a3+a4+a5+a6)+a1+a2+a3+a4
=3

When the function x belongs to [0,1], f (x) = x, the function period is 2, please write
The analytic expression of function on R is given
If 2K ≤ x ≤ 2K + 1, then 0 ≤ x-2k ≤ 1, then f (x) = x-2k (K ∈ z)
My question is why is not f (x-2k) = x-2k (x and x-2k belong to [0,)

Let x = x-2k, so f (x) = x-2k

How to prove that f (x) is a periodic function and the period is 2K from F (x + k) = - f (x)~
And there were some formulas for periodic functions in high school~

f(x)=-f(x+k)=-[-f(x+k+k)]=f(x+2k)
So the period is 2K

On periodic function
4. It is known that f (x) is an even function with period 2. When x ∈ [0,1], f (x) = x, then in the interval [- 1,3], the number of roots of the equation f (x) = KX + K + 1 (K ∈ R and k-1) about x ()
A. It is impossible to have three; B, at least one, at most four; C, at least one, at most three; D, at least two, at most four

What is k-1? Is k not equal to 1?
Anyway, the following conclusion is true
When x belongs to [- 1,0], f (x) = - X
When x belongs to [1,2], f (x) = 2-x
When x belongs to [2,3], f (x) = X-2
The pattern is sawtooth

Properties of periodic function

(1) If t (≠ 0) is a period of F (x), then - t is also a period of F (x). (2) if t (≠ 0) is a period of F (x), then NT (n is any non-zero integer) is also a period of F (x). (3) if T1 and T2 are both periods of F (x), then T1 ± T2 is also a period of F (x). (4) if f

Operational properties of periodic functions
Why is the period of F (AX + b) t / | a | if t is the period of F (x)?
Can you write down the specific operation steps?

A (x + T / |a|) + B = ax + B + T or ax + B-T
∴f[a(x+T/|a|)+b]=f(ax+b+T)=f(ax+b)
Therefore, from the definition of periodicity, we know that the period of F (AX + b) is t / | a|

The integral of periodic function in my book has a property:
The necessary and sufficient condition for ∫ (upper limit x, lower limit 0) f (x) DT to take t as period is that ∫ (upper limit t, lower limit 0) f (T) DT = 0, and ∫ (upper limit t, lower limit 0) f (x) DX = 0 is equal to ∫ (upper limit a + T, lower limit a) f (x) DX = 0?. does it not mean that the integral of all periodic functions with one cycle as upper and lower limit is zero?

Let f (x) be a periodic function and f (x) = 1, X ∈ [0,1]
The integral of F (x) over an interval of one period is equal to 1 / 2
It should be: the integrals of periodic functions in the interval of one period are equal

The proof of the properties of periodic operation of function
If t is the period of F (x), how to prove that the period of F (AX + b) is t / | a|

f(x+T)=f(x)
g(x)=f(ax+b)
g(x+T/a)=f(a(x+T/a)+b)=f(ax+b+T)=f(ax+b)=g(x)
So | T / a | = t / | a | is the period of G (x) = f (AX + b)

How to find the periodicity of periodic function!

According to the definition of the periodic function, we try to find a constant C to make the function change periodically
f(x+c)=f(x)
For example: the odd function f (x) satisfies
f(2+x)= - f(2-x)
Find the period of the function:
Because f (2 + x) = - f (2-x) = - [- f (X-2)] = f (X-2)
f(x+4)=f[(2+(x+2)]=f[(x+2)-2]=f(x)
So the function f (x) is a periodic function with period 4

Periodic properties of functions

For the function y = f (x), if there is a non-zero constant t such that when x takes every value in the domain, f (x + T) = f (x) holds, then the function y = f (x) is called a periodic function, and t is called the period of the function
If t is a period of the function, then t is an integral multiple
NT is also the period of a function. If there is a minimum positive number in all periods, the minimum positive number is called the minimum positive period