The number series problem of the second year of high school, the number series problem of the second

1、aa1 = oa1 = a * sin45°

a1a2 = oa2 = oa1 * sin45°= aa1*sin45°=a *(sin45°)^2

a2a3 = oa3 = oa2 * sin45°= a1a2*sin45°=a *(sin45°)^3

a(n+1)an = a *(sin45°)^n

a1a2a3a4… = a1a2 + a2a3 + a(n+1)an +

a*(sin45°)^2+a*(sin45°)^3+……a*(sin45°)^n+ …

lim(n->infinity)[a*(sin45°) 2*(1-(sin45°) n-1) (1-sin45°)

Note: This is a sum of proportional sequences with sin45° as the common ratio.

a (2-root number 2) is the limit when n tends to infinity.

2. Let the straight line m: 3x+4y-4=0 intersect with the x-axis at a, intersect with the y-axis at b, and ...... with the circle o1, o2, o3Delivered to A1, A2, A3 ......(lz himself marked in the diagram).

Then: by the linear equation, when x=0 => ob = 1

When y=0 => oa =4 3

So: ab=5 3

Since the circle is tangent to the straight line: O1A1 perpendicular ab = > ob = a1b =1

aa1 = ab - a1b = 2/3

So: R1 = AA1 OA = 1 2 (triangle OAB is similar to A1AO1).

R2 = (OA-1-R2) AB => R2 = 1 8 (triangular OAB is similar to A2AO2).

r3 = (oa-1-1/4-r3)/ab => r3 = 1/32

rn = [oa-2*(r1+r2+……r(n-1))-rn] ab (both are similar from triangles).

rn = (1/2)*(1/4)^(n-1)

So: sn = r1+r2+r3+......rn+……

2 3 (Finding the Limit of Proportional Sequences).

l = sn*π

Let the line m intersect with the x-axis at n and the y-axis at m

Easy to know om=1 on=4 3

mn=5/3

r1=1/2

When n 2, rn=1 4rn-1

rn is the first proportional series with an term of 1 2 and a common ratio of 1 4.

rn=（1/2）(1/4)^(n-1)

l=r1π+r2π+…rnπ+…=π/2/(1-1/4)=2π/3

a1+a3>=2 (a1*a3)=2 1=2 when the common ratio q>0.

a1+a2+a3=1+a1+a3 1+2=3 when the common ratio q<0.

a1+a2+a3=1+a1+a3≤1-2=-1

a22=a1×a3=1

a1+a2+a3=1+a1+a3 1+2 1=3 or 1-2=-1

So the conclusion is (negative infinity, 1 u 3, positive infinity).

2bn=2-2sn-2+2s=bn-b

So bn=b 3

and b1=s1=2-2s1

So b1=s1=2 3

Easy to get =(1 3) (n)*2

d=（a7-a5）/2=3

A1 = A5-4D = 2

Easy to get =-1+3n

cn=(-1+d)/3+(-1+2d）/9+(-1+3d)/27+..1+nd))*2/(3^n)

3c=(-1+2d)/3+(-1+3d）/9+(-1+4d)/27+..1+(n+1)d))*2/(3^n+1)-1+d

3c-cn=(-1+2d)/3+(-1+3d）/9+(-1+4d)/27+..1+(n+1)d))*2/(3^n+1)-[1+d)/3+(-1+2d）/9+(-1+3d)/27+..1+nd))*2/(3^n)]-1+d

d/3+d/9+d/27+..d/3^n-1+d

1*（1-1/3^n）/(1-1/3)-1+3=7/2-3*（1/3^n）/2<7/2

2.Write q as the common ratio and b as the tolerance.

a1+6d=b1*q^4

So a1=b1=6d (q 4-1).

6d/（q^4-1）+nd=6d*q^m/（q^4-1）

nd=6d*（q^(m-1)-1）/（q^4-1）

n*（q^4-1）=6*（q^(m-1)-1）

n*（q^5-q）=6*（q^m-1) *

Satisfy the above formula.

First find bn=64 (n-1), then b2=64, thus s2=1;

and a1=3, so a2=-2 that is, the tolerance of an is -5, there is a contradiction...

an+1 is also a proportional series.

then (a2+1) = (a1+1)(a3+1)a1=2

then a2=2q, a3=2q

So (2q+1) = 3(2q +1).

4q²+4q+1=6q²+3

q²-2q+1=0

q=1 so is a constant sequence.

So an=2

sn=2n