During the compulsory five number series course in senior one mathematics, the compulsory five numbe

According to f(2)=1, we get: 2 (2a+b)=1, i.e., 2=2a+b, and because f(x)=x has a unique solution: x=ax 2+bx, i.e., ax 2+(b-1)x=0 pushes out (b-1) 2-4ac=0

Since c=0, b=1 from 2=2a+b can be calculated a=1 2 from which we get the expression f(x).

Solution 1: 1: Solution a1=s1=1 2

sn=1-an

s(n-1)=1-a(n-1)

Subtract 2 formulas.

an=sn-s(n-1)=a(n-1)-anan=1/2*a(n-1)

It can be known that an is a proportional series. q=1/2

an=a1*q^(n-1)=1/2^n

2》Answer: bn=3+log4an=3+log4(1 2 n)=3+lg(1 2) n lg4=3-nlg2 (2lg2)=3-n 2

3-n/2>=0,n<=6.

Therefore, when n<=6, bn>=0

t6=|b1|+|b2|+.b6|=b1+b2+..b6=[(3-1/2)+(3-6/2)]*6/2=

When n>6, bn<0, |bn|=-bn

Therefore tn=b1+b2+.b6-(b7+b8+..bn)=2t6-(b1+..bn)=2* Solution 2: (1)an+sn=1, a(n+1)+s(n+1)=1

Subtract the two formulas: a(n+1)-an+a(n+1)=0, then an=2a(n+1), and a1=s1, then a1=1 2, so an=1 2 n

2) bn = 3 + log4an = 3-n 2 then n 6, bn 0

At n 6, bn 0

So at n 6 tn = (b1 + bn) n 2 = n (11-n) 4n 6 at t n = (n - 6) (n - 5) 4 + 15 2

The first question: from an+sn=1 to get a(n+1)+s(n+1)=1, these two formulas, and then subtract to get a(n+1)=, so the number series an is a proportional series with a common ratio of 1 2, and because a1=1 2, so an=1 2 to the n power of the second question is known by the first question,。。 Wait, is your log4an 4 base and an true number?

a1=5 6, d=-1 6, sn=-5, find n and field tremor ansn=n*a1+n(n-1)d 2

5=n*5 6+n(n-1)(-1 6) 2n=15, noisy n=-4 (discarding negative numbers).

an=a1+(n-1)d=-3/2

d=2, n=15, an=-10, find a1 and sna15=a1+14d

10=a1+14*2

a1=-38

sn=n*a1+n(n-1)d/2

15*(-38)+15*14*2 Defeat 2

Let the number of the curved macro be a1, a2, a3, and the three numbers are proportional and the product is 512, and a2 q*a2*(a2*q) can be obtained....It can be obtained that a2 3=512 a2=8 can be seen from the question: the middle term of the equal difference 2a2=a1-2+a3-2 a1 is substituted for a2 q a3 is substituted for a2q q, and q=2 is substituted for the solved a2 q into , and a1=4 a2=8 a3=16 is obtained by substituting the solved a2 q into

The topic should be problematic, otherwise the topic will be complicated.

The original question should be "the sum of the first n terms of the known difference series is sn, s7 = 49, and the middle term of the equal difference of a4 and a8 is 11".

For a: an infinite series of numbers means that the series of numbers has infinite terms, which is obviously true for b: because only the first three terms of the series are told, it is not enough to say that this is an increasing series, and the increasing series of numbers is always not less than the previous term, and not all terms are equal!

Even if 10,000 terms are given, not telling the recursive formula does not mean that it is an increasing series.

For C: a series of numbers is simply a series of numbers arranged in a certain order, and a column of function values does not indicate its order.

For D: From the above definition, you can understand!

a. An infinite sequence of numbers is an infinite number of terms, so yes.

b、1,2,3,4，……n, which is the increasing sequence. If the question is not clear that c and n are positive integers, you will read the question a few more times, and believe that you d and meet the definition of the number series, and there are recursive formulas.

Option A, this is not easy to explain, probably the ellipsis means infinity, option B, for example: 1,2,3,3,2,1,。。

Option C, not very well, is probably because the series must be in order, and the arguments corresponding to the values of a column of functions can be out of order.

Option D, which is not obvious, fits the definition of a sequence.

a5-a1=a1*(q 4-1)=15 ,a4-a2=a1*q*(q 2-1)=6 , so there is (q 4-1) (q*(q 2-1))=15 5Simplify.

2*q 2-5+2=0, q=2 or q=1 2, and bring it into a1*q=6 (q 2-1)=2 (or -8), so there is a3=a1*q*q=4 (or -4).

Is it the square of a, or is it a multiplication?

an=a(n-1)

an circular letter a(n-1)=1

Therefore, an is a series of equal shedding slow ratios with the first orange and wheel as 2 and the common ratio of 1.

an=a1q^(n-1)

an=2