Finding the Sequence Conclusion Proves Urgent !!

The recommended answer is worth it, it's good!

Obviously not.

You have to figure out what is sufficient and what is necessary.

The necessity is"x[n+1]≤x[n]" => "x1≥2"

Sufficiency is"x1≥2" => "x[n+1]≤x[n]"

You take advantage of the conditions known in necessity"x[n+1]≤x[n]"The introduction of x[n] 2 is obviously not usable in proof of sufficiency, because the known condition in sufficiency is"x1≥2", while"x[n+1]≤x[n]"It's going to be pushed out now, don't make the conditions and conclusions the opposite.

You need to understand the connotation of the proof in this question: the convergence of a series of numbers means that after n is sufficiently large (greater than n), the difference between xn and a is sufficiently small, which limits the absolute value of xn to be less than a constant after n is sufficiently large, and this constant is related to how big n is, the larger n, the smaller the deviation from a. The former finite term must have a maximum value, so that the sequence is split in two:

The first finite term is bounded, and the last infinite term is also bounded, so this sequence is bounded, and this is the meaning of taking m=max. And in fact, the bounds of the posterior infinity term here can be |a|+ any positive number, just take 1 for convenience when proving. Contradictory?

What you say is less than one is actually the upper boundary, which is the smallest of the upper bounds. 2 Of course, it is its upper boundary, note that this proof is bounded, not to find the upper boundary.

a1=2>0

Easy to prove an>0

a(n)=1 2 (a(n-1)+1 a(n-1))>=1 2*2=1 (the arithmetic mean of important inequalities is not less than the geometric mean).

an},a1=1 3,sn circle n=(2n-1)an,find the missing branch evidence: an=1 [(2n+1)(2n-1)].

Proof: Tachibana Costume Sn = n*(2n-1)an

sn+1=(n+1)*(2n+1)an+1sn+1-sn=(n+1)*(2n+1)an+1-n*(2n-1)an

an+1=(n+1)*(2n+1)an+1-n*(2n-1)an2n+3)an+1=(2n-1)an

2n+3)(2n+1)an+1=(2n-1)(2n+1)anbn=(2n-1)(2n+1)an

then bn+1=bn=.b1=1*3/3=1bn=(2n-1)(2n+1)an=1

an=1 [(2n-1)(2n+1)]Bi!!

to the conditional.

sn=4a(n-1)+2, similar can be written.

s(n-1)=4a(n-2)+2, which is obtained by subtracting the formula an=sn-s(n-1).

an=4a(n-1)-4a(n-2)

The shift term is an-2a(n-1)=2[a(n-1)-2a(n-2)], so the series is a common ratio of 2, and the first term is an equal ratio of a2-2a1, and for sn=4a(n-1)+2

Take n=2, i.e., a1+a2=4a1+2, a2=8 a2-2a1=4

an-2a(n-1)=4*2 (n-2)=2 n, i.e., an=2a(n-1)+2 n, and divide the two sides by 2 n to get an 2 n=a(n-1) 2 (n-1)+1, that is, the series is an equal difference series.

sn=4an-1+2

sn-1=4an-2+2

Subtracting the above equation yields: an=4an-1-4an-2

Shift an-2an-1=2 (an-1-2an-2) because a1=2

s2=4a1+2

So a2=8

a2-2a1=4≠0

Therefore, it is a series of equal differences.

And the general term is: an-2an-1=2 n

Divide by 2 n on both sides

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

Because a1≠0

Therefore, it is a series of equal differences with 1 as the first term and 0 as the tolerance.

, subtracting the two formulas gives a(n+1)=4an-4a(n-1).

Shifting the term gives a(n+1)-2an=2(an-2a(n-1)), so the proportionality, the common ratio is 2

2.From the above, a(n+1)-2an=(a2-2a1)*2 to the power (n-1). That is, a(n+1)-2an=2 to the (n+1) power, both sides of the equation are divided by 2 (n+1) power.

(1) For any r, t n*, there is sr st

r t), then let r=n,t=1,sn s1=n 2,sn=n 2*

a(n+1)=s(n+1)-sn=(n+1)^2a1-n^2a1=(2n+1)a1=a1+(n+1-1)(2a1)

is the first term a1, and the tolerance is a series of equal differences with d = 2a1.

2）a1=1,d=2,bn=a(b(n-1))=a1+(b(n-1)-1)d=1+2b(n-1)-2=2b(n-1)-1

So bn-1=2[b(n-1)-1].

That is, the proportional series, b(1)-1=3-1=2, q=2

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

So bn=2 n+1

tn=a1×b1+a2×b2+…+an×bn

1*3+3*5+..2n-1)*(2^n+1)

1+3+5+..2n-1+1*2+3*4+5*8+..2n-1)*2^n

n^2+1*2+3*4+5*8+..2n-1)*2^n

Let gn=1*2+3*4+5*8+.2n-1)*2^n

2gn=1*4+3*8+..2n-3)*2^n+(2n-1)*2^(n+1)

Subtract the two formulas to obtain gn=(2n-1)*2 (n+1)-2 (n+1)-2 (n)-.2^3-2

2n-1)*2^(n+1)-8*(2^(n-1)-1)-2

n-1)*2^(n+2)-2^(n+1)+6

tn=(n-1)*2^(n+2)-2^(n+1)+n^2+6

(1) Due to.

For any r, t (r, t are positive integers) have sr st=(r t) 2

So: sn s1=n 2

n∈n+)sn=n^2s1

an=sn-s(n-1)=(n^2-(n-1)^2)s1=(2n-1)a1

n n+) so, an-a(n-1) = 2a1

is a series of equal differences.

2)an=2n-1

So, bn=2b(n-1)-1

bn-1=2(b(n-1)-1)

bn-1=2^(n-1)

b1-1)=2^n

bn=2^n+1

3)tn=1*(2+1)+3*(2^2+1)+…2n-1)*(2^n+1)

1+3+……2n-1))+1*2+3*2^2+……2n-1)*2^n)

The first half is easy to calculate, equal to.

n 2, the second half of the note is s

Then 2s = 1*2 2 + 3 * 2 3 + ......2n-3)*2^n+(2n-1)*2^(n+1)

s=s-2s=1*2+2*2^2+2*2^3+……2*2^n-(2n-1)*2^(n+1)=2+8*(1-2^(n-1))/(1-2)-(2n-1)*2^(n+1)=-(2n-3)*2^(n+1)-6

s=(2n-3)*2^(n+1)+6

tn=(2n-3)*2^(n+1)+n^2+6

1.Let r=1

T is replaced by n.

It is easy to get sn=n 2·a1

sn-1=(n-1)^2·a1

Subtract to give an=(2n-1)a1

is a series of equal differences. The tolerance is 2a1

The sequence an is actually all odd numbers.

The rest is good to ask for.