Find n-> ∞lim[ 1/2a 1a2+ 1/2^2a2a3+...+ 1/2^nana(n+ 1)]
Solution:
a(n- 1)+S(n- 1)= 2(n- 1)+ 1
Two types of subtraction:
an-a(n- 1)+Sn-S(n- 1)= 2
Sn-S(n- 1)= replace:
2an=a(n- 1)+2
an =( 1/2)a(n- 1)+ 1
When n =1:a1+s1= a1+a1= 2x1+1= 3,
a 1=3/2=(4- 1)/2=(2^2- 1)/2
a2=( 1/2)a 1+ 1=( 1/2)(3/2)+ 1=3/4+ 1=7/4=(8- 1)/4=(2^3- 1)/2^2
a3=( 1/2)a2+ 1=( 1/2)(7/4)+ 1=7/8+ 1= 15/8=( 16- 1)/8=(2^4- 1)/2^3
a4=( 1/2)a3+ 1=( 1/2)( 15/8)+ 1= 15/ 16+ 1=3 1/ 16=(32- 1)/ 16=(2^5- 1)/2^4
Guess what:
ak=(2^(k+ 1)- 1)/2^k
a(k+ 1)=( 1/2)ak+ 1=( 1/2)[(2^(k+ 1)- 1)/2^k]+ 1=(2^(k+ 1)- 1)/2^(k+ 1
)+ 1=[2^(k+ 1)- 1+2^(k+ 1)]/2^(k+ 1)=[2x2^(k+ 1)- 1]/2^(k+ 1)
=[2^(k+2)- 1]/2^(k+ 1)
The guess is correct.
1/2^kaka(k+ 1)=( 1/2^k)[2^k/(2^(k+ 1)- 1)][2^(k+ 1)/(2^(k+2)- 1)]
=2^(k+ 1)/(2^(k+ 1)- 1)(2^(k+2)- 1)
=2^(k+ 1)(2- 1)/(2^(k+ 1)- 1)(2^(k+2)- 1)
=[2^(k+2)-2^(k+ 1)]/(2^(k+ 1)- 1)(2^(k+2)- 1)
=[(2^(k+2)- 1)-(2^(k+ 1)- 1)]/(2^(k+ 1)- 1)(2^(k+2)- 1)
= 1/(2^(k+ 1)- 1)- 1/(2^(k+2)- 1)
∴
1/2a 1a2+ 1/2^2a2a3+...+ 1/2^nana(n+ 1)
=( 1/3- 1/7)+( 1/7- 1/ 15)+( 1/ 15- 1/3 1)+...+[ 1/(2^(n+ 1)- 1)- 1/(2^
(n+2)- 1)]
= 1/3- 1/(2^(n+2)- 1)
n->； ∞, the latter term of the above formula approaches 0, and the above formula approaches 1/3.
2.an, a(n+ 1) is the root of equation 4nx 2-4nbnx+1= 0, and the sum of all terms in infinite sequence {bn} is
1 1/3, find the sum of all the items in the infinite series {an}.
Solution:
According to Vieta's theorem:
ana(n+ 1)= 1/4^n
an+a(n+ 1)=4^nbn/4^n=bn
The latter item:
a(n+ 1)a(n+2)= 1/4^(n+ 1)
Department:
a(n+2)/an= 1/4
Therefore, the odd and even terms of the sequence {an} are geometric series with two common ratios 1/4 respectively. All projects
sum = a 1/( 1- 1/4)+A2/( 1- 1/4)=(4/3)(a 1+A2)=(4/3)。
Re-learn {bn}
b 1=a 1+a2
b2=a2+a3
B3 = a3+a4 = a 1/4+a2/4 =(a 1+a2)/4 = b 1/4
B4 = a4+a5 = a2/4+a3/4 =(a2+a3)/4 = B2/4
bn = an+a(n+ 1)= a(n-2)/4+a(n- 1)/4 =[a(n-2)+a(n- 1)]/4 = b(n-2)/4
So {bn} is also an odd term, and the even term is a geometric series. The two commonly used ratios are 1/4 respectively. Sum of all items
= b 1/( 1- 1/4)+B2/( 1- 1/4)=(4/3)(b 1+B2)= 1 1/3
b 1+B2 = 1 1/4 = a 1+a2+a2+a3 = a 1+2 a2+a3
a 1a2= 1/4，a 1= 1/4a2
a2a3= 1/4^2,a3= 1/ 16a2
Substitution:
1/4a 2+2 a2+ 1/ 16 a2 = 1 1/4
2 a2+5/ 16 a2- 1 1/4 = 0
32a2^2-44a2+5=0
(8a2- 1)(4a2-5)=0
A2= 1/8, or a2=5/4.
A 1= 1/4a2=2 or a1=1/4a2 =1/5.
B1= a1+a2 = 2+1/8 =17/8, or B 1 = 1/5+4 = 29/20.
A3 =116a2 =1/2 or a3 =1/6a2 =1/20.
B2=a2+a3=5/8, or B2 = A2+A3 = 5/4+1/20 = 26/20 =13/10.
B1+B2 =17/8+5/8 = 22/8 =1/4, or b1+B2 = 29/20+26/20 = 55.
correct
Sum of {an} items = (4/3) b1= (4/3) (17/8) =17/6.
or
The sum of {an} items = (4/3) b1= (4/3) (29/20) = 29/15.
3. n∈N*, y = n (n+1) x 2 and y=(2n+ 1)x- 1, the projection length of the line segment on the x axis.
And ...
Solution:
Replaced:
n(n+ 1)x^2=(2n+ 1)x- 1
n(n+ 1)x^2-(2n+ 1)x+ 1=0
(NX- 1)[(n+ 1)x- 1]= 0
x 1= 1/(n+ 1)，x2= 1/n
| x 1x 2 | = x2-x 1 = 1/n- 1/(n+ 1)
Total number of projects =
( 1- 1/2)+( 1/2- 1/3)+( 1/3- 1/4)+...+( 1/n- 1/(n+ 1))
= 1- 1/(n+ 1)
n->； ∞, the above formula approaches 1.
4. The sum of the first n items in the sequence {an} Sn = Na+N (n- 1) b, (n = 1, 2, ...), A and B are constants.
(1) Proof: {an} is arithmetic progression.
(2) It is proved that the points Pn(n= 1, 2, ...) all fall on the same straight line with (an, Sn/n- 1) as the coordinates.
And write the equation of this straight line.
solve
( 1)
Prove:
a 1=S 1=a
s(n- 1)=(n- 1)a+(n- 1)(n-2)b
an = Sn-S(n- 1)= a+(n- 1)2b = a 1+(n- 1)2b
This is a arithmetic progression with an error of 2b.
(2)
Serial number/n-1= a+(n-1) b-1
Pn(a+(n- 1)2b，a+(n- 1)b- 1)
P(n+ 1)(a+2nb，a+nb- 1)
PNP (n+1) = [(a+nb-1)-(a+(n-1) b-1)]/[(a+2nb)-(a+(n-65433)
=b/2b= 1/2= fixed value
Therefore, P 1, P2, ..., Pn***
P 1(a，a- 1)，
y=( 1/2)(x-a)+(a- 1)
Replacement of Pn coordinates:
Left =a+(n- 1)b- 1
right =( 1/2)(a+(n- 1)2 b-a)+(a- 1)
=( 1/2)((n- 1)2b)+(a- 1)
=(n- 1)b+(a- 1)
=a+(n- 1)b- 1
Left = right, correct.