If f (x) = -- cos3.14 * x, X is greater than 0, f (x + 1) + 1, X is less than or equal to 0, then the value of F (4 / 3) + F (- 4 / 3)

If f (x) = -- cos3.14 * x, X is greater than 0, f (x + 1) + 1, X is less than or equal to 0, then the value of F (4 / 3) + F (- 4 / 3)

Building lord, 3.14 I write π directly, OK
f(4/3)+f(-4/3)
＝-cos（4π/3）＋f(-1/3)+1
＝cos(π/3)＋f（2/3）+2
＝cos(π/3)-cos（2π/3）+2
＝1+1+2
＝4

Given f (x) = {cos π x, X ＜ 1 {f (x-1), X ＞ 1, find the value of F (1 / 3) + F (4 / 3)
It's a process

1/31
Then f (4 / 3) = f (4 / 3-1) = f (1 / 3) = 1 / 2
So the original formula is 1

Given that the minimum value of the function f (x) = 2Sin ω x (ω > 0) in the interval [− π 3, π 4] is - 2, then the minimum value of ω is equal to ()
A. 23B. 32C. 2D. 3

If the minimum value of the function f (x) = 2Sin ω x (ω > 0) in the interval [− π 3, π 4] is - 2, then the value range of ω x is [− ω π 3, ω π 4], the minimum value of − ω π 3 ≤− π 2 or ω π 4 ≥ 3 π 2 is equal to 32, so B

Ask a question about the sequence of higher one, A1 = 1, a (n + 1) = 2An + 2 ^ n, find an,

A (n + 1) = 2An + 2 ^ n divided by 2 ^ n + 1a (n + 1) / (2 ^ n + 1) = an / (2 ^ n) - 1A (n + 1) / (2 ^ n + 1) - an / (2 ^ n) = 1 at the same time, so {an / (2 ^ n)} is an arithmetic sequence with 1 / 2 as the first term and 1 as the tolerance, so an / (2 ^ n) = 1 / 2 + (n-1) * (1 / 2) = n / 2, so an = (n / 2) * (2 ^ n) = n * 2 ^ (n-1)

The sequence an satisfies A1 = 1, A2 = 3 / 2, an + 2 = 3 / 2An + 1-1 / 2An, and N belongs to positive integer (n + 2 and N + 1 are diagonal markers)
(1) Note DN = an + 1-an to prove DN as an equal ratio sequence
(2) The general term formula of the sequence an
(3) Let BN = 3n-2 find the first n terms and Sn of sequence {an * BN}

an+2-an+1=1/2(an+1-an)
dn+1=dn/2
dn+1/dn=1/2
{DN} is an equal ratio sequence with a common ratio of 1 / 2
d1=1/2
The total of the first n items of {DN} is SN
Sn=1-1/2^n
Sn=a2-a1+a3-a2+a4-a3…… an+1-an=an+1-a1=an+1-1
1-1/2^n=an+1-1
an+1=2-1/2^n
an=2-1/2^(n-1)
Let CN = an * BN
cn=(3n-2)-(3n-2)/2^(n-1)
Let Sn = S1-S2
S1=3(1+2+3+…… n)-2n=(3n^2-n)/2
S2=1/2^0+4/2^1+7/2^2…… (3n-2)/2^(n-1)
S2/2=1/2^1+4/2^2+7/2^3…… (3n-2)/2^n
Front minus back
S2/2=1/2^0+3/2^1+3/2^2…… 3/2^(n-1)-(3n-2)/2^n
=1-(3n-2)/2^n+3[1/2+1/2^2+…… 1/2^(n-1)]
=4-3/2^(n-1)-(3n-2)/2^n
S2=8-3/2^(n-2)-(3n-2)/2^(n-1)
Sn=S1-S2=(3n^2-n)/2-8+3/2^(n-2)+(3n-2)/2^(n-1)

In the sequence {an}, A1 = 2, 2An + 1-2an = 1, then the value of A101 is ()
A. 49B. 50C. 51D. 52

In the sequence {an}, A1 = 2, from 2An + 1-2an = 1, an + 1 − an = 12 is obtained. The sequence {an} is an arithmetic sequence with the first term of 2 and the tolerance of 12, A101 = 2 + 100 × 12 = 52