The math problem of the first year of high school, anxious, come and see

Because (an,an+1) is on the image of the function f(x)=x 2+2x.

a(n+1)=an^2+2an

1+a(n+1)=an^2+2an +1=(1+an)^2

LG3(1+AN+1)=2LG3(1+AN),(N>=1), so it is proportional.

Write bn=(1 an)+[1 a(n+2)], and find the first n terms and sn. of the sequence

b1=(1/a1)+[1/a3]=1/2+1/80=41/80,b2=(1/a2)+[1/a4]=1/8+1/6560=83/6560,b3=(1/a3)+[1/a5],.bn = (1 an) + [1 a(n+2)], the first n terms of the series and sn

sn=b1+b2+b3+..bn=

1/a1+1/a3)+(1/a2+1/a4)+(1/a3+1/a5)+.1/an+1/a(n+2)]=

.Because 1 [(a1)*(a1+2)]=(1 2)[1 (a1)-1 (a3)],1 [(a2)*(a2+2)]=(1 2)[1 (a2)-1 (a4)],1 [(a3)*(a3+2)]=(1 2)[1 (a3)-1 (a5)],1 [(a(n-1))*a(n-1)+2)]=(1 2)[1 (a(n-1)-1 (a(n+1))]1 [(an))*an+2)]=(1 2)[1 (an)-1 (a(n+2))]1 [(a(n+1))*a(n+1)+2)]=(1 2)[1 (a(n+1)-1 (a(n+3))]So, sn=1 a1+(1 2)[1 (a2)-1 (a4)]+1 2)[1 (a1)-1 (a3)]+

1/2）[1/(a3)-1/(a5)]+1/2）[1/(a(n-1)-1/(a(n+1))]

1/2）[1/(a(n+1)-1/(a(n+1))]=

3/2)(1/a1)+(1/2)(1/a2)-(1/2)[1/(a(n+3))]=

3/2)(1/2)+(1/2)(1/8)-(1/2)[1/(a(n+3))]=

13/16-(1/2)[1/(a(n+3))]

This is an adaptation of the finale of the 06 Shandong volume, but your title is very difficult to understand, and it took me a lot of effort to understand the meaning ...... you expressed

a(n+1)=an^2+2an

a(n+1)+1=an^2+2an+1=(an+1)^2b(n+1)/bn=log3(1+a(n+1))/log3(1+an)=2

Therefore, it is a proportional series.

a1+1=3

b1=1bn=2^(n-1)

an=3^(2^(n-1))-1

1 (an+2)=(1 an)-[2 a(n+1)] is a good proof that a(n+1)+1=xn

Apparently there is. 1 (xn+1)=1 (xn-1)-2 (xn 2-1).

This question seems to be problematic ......At least you don't express it very clearly.

o(0,0), which is calculated from the formula for the distance between two pointsoa|，|ob|，|ab|values of 5, 2, 5. According to the cosine formula, it can be calculated that the cosine value of the angle AOB is 2 10

Using the cosine theorem, the final result is (2) (10).

The cosine theorem will do it. This is the content of the second year of high school.

（1）ac=asinθ，ab=acosθ，s1=ab×ac/2=(a^2/4)sin2θ.

Make ad bc in d, cross pq in e, let the side length of the square be x, ad=absin =(asin2) 2, ps:bc=ae:ad

x/a=[(asin2θ)/2-x)]/(asin2θ/2),∴x=asin2θ/(2+sin2θ),s2=x^2=(asin2θ)^2/(2+sin2θ)^2(2)

When a is fixed and changes, s1 s2 = (2 + sin2 ) 2 4 sin 22 , let 1 sin2 = t, 0 s1 s2 = t 2 + t + 1 + 1 4 1 + 1 1 4 i.e., when a is fixed and is 45 °, s1 s2 reaches the minimum value of 9 4

f(2x+1) is an odd function.

That is, the center of symmetry is (0,0).

Move him 1 2 units to the right.

Even if f[2(x-1 2)+1]=f(2x), then the center of symmetry will also be shifted to the right by 1 2 units.

Yes (1 2, 0) choose c

Solution: Easy to find: a=, 1, if a b=, then.

2 +4(a+1)+(a-5)=0, a +4a+3=0 gives a=-1 or -3

2. If a b = a, then a contains b, and there are two cases.

1) b is an empty set, at this time [2(a+1)] 4(a -5) 0, 2a+6 0 group digging min a -3

2) b is not an empty set, i.e. there are one or two elements in b.

When there is an element in b, [2(a+1)] 4(a -5)=0, the solution is a=-3, and this scattered form is b=, which is in line with the title.

When there are two elements in the trillion branch b, a -3, then b = a = according to Veda's theorem, get.

2(a+1)=1+2, and (a-5)=1 2 has no solution at this time. To sum up, the value range of a is (+3] Thank you.

1) Positive and negative root number three 2) I don't understand the meaning of the title, how can it be the range!

1.Bringing Brother Hui x=2 into B gives 2 +4(A+1)+(A -5)=04+4A+4+A -5=0

a²+4a+3=0

a+3)(a+1)=0

Solution: a=-1 or -3

a ub=a

b = or or empty set, x 2 + 2 (Hui Ridge A + 1) x + (A 2-5) = 0 pre-solution dust infiltration is two identical solutions or no solution.

4(a+1)²-4（a²-5）=4a²+4a+1-4a²+20=4a+21≤0

a≤-21//4

Let the sets a=,b=,1If a b=, find the real value a.

Because 2 is the solution of b, substitute 2 into have.

4+4(a+1)+(a^2-5)=0

a^2+4a=0

a = 0 or -4

2.If a ub=a, find the range of the real number a.

a ub=a, i.e. b is a subset of a.

Then first a=

Therefore, the solution of b can only be in 1, old knowledge 2.

The situation of 2 has been given above, and 1 substitution has it.

1＋2（a＋1）+(a^2-5)=0

a^2＋2a－3＝0

a 3 or 1

Therefore, the return generation A test is counted, and those who are lacking are included.

Found A1 and 4 can.

Since f(x+y)=f(x)+f(y), f(1 3+0)=f(1 3)+f(0), substituting f(1 3) gives f(0)=0To judge its parity, you can do this, f(0) = f(x+(-x)) = f(x) + f(-x), you can get f(x) = -f(-x). This is an odd function.

Third, f(x)+f(2+x)=f(x+2+x)<2, f(2+2x)<2Because 2>0, according to the title, f(2+2x)-2<0, f(2+2x)+(2)<0, since it is an odd function, 2f(1 3)=2, -2f(1 3)=f(-1 3)+f(-1 3), therefore, f(2+2x)+f(-1 3)+f(-1 3)<0, get, f(2+2x-1 3 3-1 3)<0, according to the title, 2+2x-2 3>0, get x>-4 6

1.Let x=0 and y=0.

f(0)=f(0)+f(0)=2f(0), so f(0)=02Let x=-y, then f(0)=f(x)+f(-x)=0, so f(x)=-f(-x) odd.

f(2/3)=2

then x+2+x<2 3

x<-2/3

Write -1 as log(a) 1 a, 1 as log(a)a log(a) 1 a log(a)2 3 log(a)a (the base of the formula is a).

When the base a 1 is an increasing function, 1 a 2 3 a is solved to get a 3 2

When the base number 0 a 1 is a shear function, a 2 3 1 a is solved to obtain 0 a 2 3

Range of a: 0 a 2 3 or a 3 2

-11, the function loga(x) is the increment function, so.

then from the inequality gives a>(3 2);

If 0 (2 3), then 0

Write -1 as log(a) 1 a, and 1 as log(a)a, then log(a) 1 a log(a)2 3 log(a)a (the base of the equation is a).

When a 1, is an increasing function, 1 a 2 3 a gives a 3 2

When 0 a 1, it is a subtraction function, a 2 3 1 a so 0 a 2 3

Range of a: 0 a 2 3 or a 3 2