The general term formula of sequence 1,111

The general term formula of sequence 1,111

Look first
10,100,1000.
The general term is an = 10 ^ n
9,99,999,9999
The general term is
an=10^n-1
∴1,11,111,1111
The general term is
an=(10^n-1)/9
How to write the general formula of sequence 1, 111, 111
Such as the title
a(n)=((10^n)-1)/9
Where n = 1,2,3,4,5
Let f (β) = (2cos ^ 3 β + sin (2 Π - β) + sin (Π / 2 + β) - 3) / (2 + 2cos ^ 2 (Π + β) + cos (- β)), find the value of F (Π / 3)
f(∏/3)=[2cos^3 ∏/3＋sin(2∏-∏/3)+sin（∏/2＋∏/3）-3]/[2+2cos^2(∏+∏/3)+cos(-∏/3)]=[2*(1/2)^3＋sin(-∏/3)+sin（5∏/6)-3]/[2+2cos^2(4∏/3)+1/2]=(1/4-√3/2+1/2-3)/[2+2*(-1/2)^2+1/2]=(-√3/2-9/4)/3=...
Predigestion f (β): F (β): F (β): F (β): F (β)
f(β)=(2(cosβ)³＋sin(2π-β)+sin(π/2＋β)-3)/(2+2(cos(π+β))²+cos(-β))
=(2(cosβ)³-sinβ+cosβ-3)/(2+2(cosβ)²+cosβ)
=(2 (COS β) & sup3; - 3) / (2 + 2 (COS β) & sup2... Expansion
Predigestion f (β): F (β): F (β): F (β): F (β)
f(β)=(2(cosβ)³＋sin(2π-β)+sin(π/2＋β)-3)/(2+2(cos(π+β))²+cos(-β))
=(2(cosβ)³-sinβ+cosβ-3)/(2+2(cosβ)²+cosβ)
=(2(cosβ)³-3)/(2+2(cosβ)²+cosβ)
Then find the value of F (π / 3)
f(π/3)=(2(cos(π/3))³-3)/(2+2(cos(π/3))²+cos(π/3))
=(2(1/2)³-3)/(2+2(1/2)²+1/2)
=-11 / 12. Put away
The monotonicity of function f (x) = ax1-x2 (a ≠ 0) on interval (- 1,1) is discussed
F ′ (x) = a (x2 + 1) (1-x2) 2; F ′ (x) > 0 when a > 0; f (x) increases monotonically on (- 1,1); F ′ (x) < 0 when a < 0; f (x) decreases monotonically on (- 1,1)
Given the solution set [- 1, 3] of inequality AX2 + BX + C ≥ 0, then the monotone increasing interval of function f (x) = − 16bx3 + AX2 + CX + m is ()
A. （-∞，-1），（3，+∞）B. （-1，3）C. （-3，1）D. （-∞，-3），（1，+∞）
∵ the solution set [- 1, 3] of inequality AX2 + BX + C ≥ 0, then B = − 2Ac = − 3A & nbsp; a < 0 & nbsp; ∵ function f (x) = − 16bx3 + AX2 + CX + m, ∵ f '(x) = - 12bx2 + 2aX + C = AX2 + 2ax-3a = a (x-1) (x + 3), Let f' (x) > 0, the solution is - 3 < x < 1, ∵ function f (x) = − 16bx3 + AX2 + CX + m, the monotone increasing interval is: (- 3, 1), so the answer is: C
We know that an + Sn = n.1, let BN = an-1, and prove that {BN} is an equal ratio sequence
Sn + an = ns (n-1) + a (n-1) = n-1an + an-a (n-1) = 12An = a (n-1) + 1bn = an-12an-2 = a (n-1) - 12bn = B (n-1) BN = (1 / 2) B (n-1), so the ratio a1 + A1 = 1A1 / 2B1 = A1-1 = - 1 / 2bn = (1 / 2) ^ (n-1) * B1 = - (1 / 2) ^ n = an-1an = 1 - (1 / 2) ^ n
F () θ = sin ^ (2 π - θ) + sin (π / 2 + θ) - 3 / 2 + 2cos ^ 2 (π + θ) + cos (- θ) to find f (π / 3)
f(θ)=sin^2(2π-θ)+sin(π/2+θ)-3/2+2cos^2(π+θ)+cos(-θ）
=sin^2θ+cosθ-3/2+2cos^2θ+cosθ
=cos^2θ+2cosθ-1/2
f(π/3)=1/4+√3-1/2=√3-1/4
Judging the monotonicity of function f (x) = ax / (x2 + 1) (a ≠ 0) on interval (- 1,1)
Let g (x) = 1 / F (x) = 1 / a [x + 1 / x] and y = x + 1 / X decrease on (- 1,0) and increase on (0,1). When a > 0, then G (x) = 1 / F (x) = 1 / a [x + 1 / x] decrease on (- 1,0], increase on (0,1), then f (x) = 1 / g (x) = ax / (x2 + 1), increase on (- 1,0] and decrease on (0,1)
Given the function f (x) = x | X-2 | (1) write the monotone interval of F (x); (2) solve the inequality f (x) < 3
（1）∵f（x）=x|x-2|=x2−2x  (x ≥ 2) − x2 + 2x (x < 2), f (x) increases monotonically on (- ∞, 1], [2, + ∞) and decreases monotonically on [1,2]; (2) if x ≥ 2, f (x) < 3 ⇔ x2-2x < 3, the solution is 2 ≤ x < 3; if x < 2, f (x) < 3 ⇔ - x2 + 2x < 3, that is, x2-2x + 3 = (x-1) 2 + 2 > 0, x < 2 satisfies the problem In conclusion, the solution set of inequality f (x) < 3 is {x | x < 3}
1. Let the common ratio of the equal ratio sequence {an} be q, and the sum of the first n terms be Sn > 0 (n = 1,2,3,...) ）(1) The range of Q
(2) Let BN = a subscript (n + 2) - 1.5A subscript (n + 1), and note that the sum of the first n terms of {BN} is TN. try to compare the size of Sn and TN
A (n) = AQ ^ (n-1), a = a (1) = s (1) > 0, when q = 1, s (n) = Na > 0. Meet the requirements. When q is not equal to 1, s (n) = a [Q ^ n-1] / (Q-1). When Q > 1, Q ^ n-1 > 0, Q-1 > 0, s (n) = a [Q ^ n-1] / (Q-1) > 0. Meet the requirements. - 1-1. B (n) = a (n + 2) - 1.5A (n + 1) = AQ ^ (n + 1) - 1