Help me solve a high school number series problem 20

Question 1: sn=a1n+n(n-1)d 2 s2n=2na1+n(2n-1)d

From s2n-2sn=n 2, we get d=1, which is carefully calculated, so an=1+(n-1)*1=n, so an=n

Second question: Because an=n so bn=n+q n so tn=b1+b2+b3....bn=(1+q)+(2+q^2)(3+q^3)..

n+q^n)=(1+2+3...n)+(q+q^2+q^3...q^n)=n(n+1)/2+n

s2=1+a2

s2-2s1=2² (1+a2)-2*1=4 a2=7

d=a2-a1=7-1=6

Because it is a series of equal differences, an=a1+(n-1)d=1+6*(n-1)=6n-5

bn=n+q^an=n+q^(6n-5)

Divide the bn into two parts: n and q (6n-5).

The first n terms and tn of bn are divided into two parts: mn=n (the sum of the first n terms of the equal difference series) and kn=q (6n-5) (the sum of the first n terms of the equal ratio series).

Knowing that mn is an equal difference series with the first term 1 and a tolerance of 1, the sum of its first n terms is n(n+1) 2

kn is a proportional sequence with the first term q and the tolerance q 6, and the sum of its first n terms is [q-q (6n-4)] (1-q) (q≠1) If q=1, it is a constant series with a constant of 1, and the sum of its first n terms is n

So tn=n(n+1) 2+[q-q (6n-4)] (1-q).

or tn=n(n+1) 2+n

s2=1+a2

s2-2s1=2² (1+a2)-2*1=4 a2=7d=a2-a1=7-1=6

Knowing that mn is an equal difference series with the first term 1 and a tolerance of 1, the sum of its first n terms is n(n+1) 2

kn is a proportional sequence with the first term q and the tolerance q 6, and the sum of its first n terms is [q-q (6n-4)] (1-q) (q≠1) If q=1, it is a constant series with a constant of 1, and the sum of its first n terms is n

So tn=n(n+1) 2+[q-q (6n-4)] (1-q) or tn=n(n+1) 2+n

where the tolerance is found first: d=(a4-a1) 3=(2-8) 3=-2;

Then set the formula an=a1+(n-1)d=8+(n-1)*(2)=10-2n

The general formula is an=10-2n

sn=(a1+an)*n/2=(8+10-2n)*n/2=9n-n^2

(1)sn+1=3/2sn +1 2sn+1=3sn +2 , 2sn+1 +4=3sn +6 , 2(sn+1 +2)=3(sn +2)

then, sn +2 is the proportional sequence of 3 with the first term being 3 and the common ratio being 3 2. so sn=3(3/2)^n-1 。

2) 1 an=(2 3) n-1 tn=3{1-(2 3) n) sn=3(3 2) n-1 -2

I'll just talk about the second question.

If there are three terms in , they can form a series of equal differences, then there is 2an=(an-1)+(an+1).

That is, 2*(3*2 n-3)=3*2 (n+1)-3+3*2 (n-1)-3,3*2 (n+1)-6=3*2 (n+1)+3*2 (n-1)-6

3*2 (n-1)=0, i.e. 2 (n+1)=0, however, this is not possible, there are no terms in the sequence, so that they can form an equal difference series.

Using an=sn-s(n-1), the recursive general term of question (1) can be obtained.

Is it a mathematical induction method to prove that it doesn't exist??

Haha, I'm also a freshman in high school, I'll give it a try.

bn=2 (an) bn+1 = 2 [a(n+1)], then bn+1 bn (quotient between the last term and the former term of the number bn) =2 [a(n+1)-an]=2 (3n-2-3n+5)=2 3=8

For the first proportional sequence b1=2 a1, then the sum is also the sum of the proportional sequence, and the summing formula is OK.

And then the second way:

Is that most positive period the smallest positive period?

If so, use the cosx double angle formula y=cos(2x) then the minimum positive period is t=2 2= .

I don't know if it's right or not, I hope the landlord uses pulling

Both sides of the equation are divided by a(n)a(n+1).

2 n a(n+1)-2 (n-1) a(n)=1 The resulting series is a series of equal differences with the first term 1 and a tolerance of 1.

Then 2 (n-1) a(n) = n

So a(n)=2 (n-1) n

Because 2 nan=2 (n-1)a(n+1)+ana(n+1) is divided by ana(n+1) on both sides at the same time

2 n a(n+1)=2 (n-1) an+1, so let 2 (n-1) an=bn

So b(n+1)-bn=1 bn=n

So a(n)=2 (n-1) n

When n=n-1, the formula is obtained, and the formula obtained is obtained by subtracting the original formula. Getting the relationship between an and an-1 terms, it is derived.

Solution: The sum of the first n terms of the series is s[n], na[n+1]=s[n]+n(n+1).

ns[n+1]-ns[n]=s[n]+n(n+1)ns[n+1]-(n+1)s[n]=n(n+1)s[n+1]/(n+1)-s[n]/n=1a[1]=2

s[1]=a[1]=2

It is a series of equal differences with the first term s[1] 1=2 and a tolerance of 1.

That is: s[n] n=2+(n-1)=n+1

s[n]=n(n+1)

s[n-1]=(n-1)n

Subtract the above two equations to get:

a[n]=2n

a1=2na(n+1)=sn+n(n+1), n-1)an=s(n-1)+(n-1)n, subtract the two equations to get na(n+1)-(n-1)an=an+2n, so na(n+1)-nan=2n

So a(n+1)-an=2

This results in a series of equal differences, with the first term being 2 and the tolerance being 2, an=2n

When b=2an=sn-sn-1=2(an-an-1)+2 (n-1)an=2(an-1-2 (n-1)

With the construction method. an-2 (n-1)=2[an-1-2 (n-1)][an-2 (n-1)] an-1-2 (n-1)]=2, so it is a proportional series.

If the problem is wrong, let x=0, then tanx=0, and the original equation is not true.

Simple. The meaning is roughly as follows: let k = tanx, find the two values of k, the smaller one is k1, the larger one is k2, because the arctangent is a periodic function of , an is arctan k1 + arctan k2 + arctan k1+

Arctan K2++ up to the sum of the nth one, that's all.