If the sequence {an} satisfies A1 = 3 / 2, an + 1 = an ^ 2-An + 1 (n belongs to positive integer), then M = 1 / A1 + 1 / A2 + +What is the integer part of 1 / A2009?

It is known from the title that a (n + 1) - 1 = a (n) * (a (n) - 1), 1 / (a (n + 1) - 1) = 1 / [a (n) * (a (n) - 1) = 1 / (a (n) - 1) - 1 / a (n); 1 / (a (n) - 1) - 1 / (a (n + 1) - 1) = 1 / a (n), and 1 / A1 + 1 / A2 + is obtained by the method of accumulation +1 / A2009 = 1 / (A1-1) - 1 / (a2010-1) = 2-1 / (a2010-1)

If the sequence {an} satisfies A1 = 3 / 2, an + 1 = an2-an + 1, then M = 1 / A1 + 1 / A2 + +The integer part of 1 / A2010 is
If the sequence {an} satisfies A1 = 3 / 2, an + 1 = the square of an - an + then M = 1 / A1 + 1 / A2 + +The integer part of 1 / A2010 is

If the sequence {an} satisfies A1 = 0.5, an = 1-1 / (an-1), and N is greater than or equal to 2, then what is A2009 + A2010 equal to

a1=1/2
So A2 = 1-2 = - 1
a3=1+1=2
a4=1-1/2=1/2=a1
So three in one cycle
2009 / 3 remainder 2
So A2009 = A2
a2010=a3
So the original formula = - 1 + 2 = 1

Given that the quadratic function f (x) = ax ^ 2 + BX + C satisfies f (0) = f (1 / 2), and the minimum value of F (x) is - 1 / 8, let the sequence {an} be forward n-direction

(1) It is known from the title: {a + B = 0A ＞ 0-b24a = - 18, the solution is {a = 12b = - 12,
So f (x) = 12x2-12x (4 points)
(2) TN = a1a2an = (45) n2-n2, (5 points)
TN-1 = a1a2an-1 = (45) (n-1) 2 - (n-1) 2 (n ≥ 2) (7 points)
Ψ an = tntn-1 = (45) n-1 (n ≥ 2), (9 points)
So an = (45) n-1 (n ∈ n *) (10 points)
TN = (45) 0 + 2 (45) 1 + 3 (45) 2 + + n (45) n-1,45tn = 45 + 2 (45) 2 + + (n-1) (45) n-1 + n (45) n (11 points)
15tn = 1 + 45 + (45) 2 + + (45) n-1-n (45) n, (13 points)
15tn = 1 - (45) n1-45-n (45) n, TN = 25 - (25 + n) (45) n, (15 points)

Given that the quadratic function f (x) = ax ^ 2 + BX satisfies the condition 1: F (0) = f (- 1) 2: the minimum value of F (x) is - 1 / 8, find the analytic expression of function f (x), let the sequence {an}
The product of the first n terms is TN, and TN = (2 / 3) ^ f (n)

f（x）=0.5x^2+0.5x

For example, s (134) = 1 + 3 = 4, e (134) = 4. S (1) + s (2) + +S（100）= ．E （1）+E（2）+… +E（100）= ．

The times of 1, 3, 5, 7 and 9 in 1-99 are the same, 20 times in total, s (1) + s (2) + s (3) + +S (100) = (1 + 3 + 5 + 7 + 9) × 20 + 1 = 5011 ~ 99, the times of 2, 4, 5, 8 are the same, a total of 20 times e (1) + e (2) + e (3) + +E (100) = (2 + 4 + 6 + 8) × 20 = 400, so the answer is 501; 400

Among the digits of natural number n, the sum of odd digits is s (n), and the sum of even digits is e (n). For example, s (145) = 6, e (145) = 4, S1 + +S(100)=
Similarly, E1 + +E（100）=

First, S100 = 1, S100 = 0;
Then 1-9 of the 99 numbers from 1 to 99 appear 10 times on the individual position and 10 times on the ten position
And 1 + 2 + 3 +. + 9 = 45
So S1 +. S100 = S1 + S2 +. S99 + S100 = 45 * 10 + 1 = 451,
E1+.E100=E1+E2+.E99+E100=45*10+0=450

If the sequence {an} satisfies A1 = 3 / 2, an + 1 = an ^ 2-An + 1 (n belongs to positive integer), then M = 1 / A1 + 1 / A2 + +What is the integer part of 1 / A2009?