A number series problem for a detailed explanation, thank you

f2007(11)=f1(f2006(11))=f1(f1(f2005(11)))=f1(f1(f1(f2004(11)))

.=f1(f1(f1 ..f1(11)..Altogether 2007 layers f1 (f1 (f1 .f1(4)..=f1(f1(f1 ..f1(16)..

f1(f1(f1 ..f1(49)..=f1(f1(f1 ..f1(169)..Inside-out 5th floor.

f1(f1(f1 ..f1(256)..6 layers from the inside to the outside f1 (f1(f1 ..)

f1(169)..From the inside out, layer 7 f1 (f1 (f1 .f1(256)..

Inside-out layer 8 f1 (f1 (f1 .f1(169)..9th layer F1 (F1(256)) from the inside out, 2006 layer F1 (169) from the inside out, 2007 layer from the inside out.

It's simple, f2007(11)=f2004(f3(11))=f2004(49)=f2003(169)=f2002(256).

After the occurrence of 256, the loop starts (in fact, it should be said that it starts after the appearance of 169), and after the odd number of f operations, it is (169) and the even number of times (256) is easy to verify.

1+3+5+ （2n-1）

You can do this by analogy: For example, 1+3+5 is (1+5) 2=31+3+5+7 is (1+7) 2=4

So adding to (2n-1) is (1+2n-1) 2=n

Numerator: The sum is n(n+1) 2

Denominator: The sum is n

After about a minute, the left is.

n+1）/（2n）

n+1）/（2n）=10/19

Solution: n=19

Sequences and functions, as well as observation and coincidence techniques.

a2012-1)^3+2014a2012=0

a3)^3-3(a3)^2+2017a3

a3)^3-3(a3)^2+2014a3+3a3

a3-1)^3+2014a3+1=4029

So(a3-1) 3+2014a3=4028

Let f(x)=(x-1) 3+2014x,f'(x)=3(x-1) 2+2014 0

So f(x) increases monotonically, f(a3)=2048 0=f(a2012).

So A3 A2012

Get: (a3-1) 3+2014a3+(a2012-1) 3+2014a2012=4028

a3+a2012-2)[(a3-1)^2-(a3-1)(a2012-1)+(a2012-1)^2]+2014(a3+a2012)=4028

Let a3+a2012=t,g(t)=[a3-1) 2-(a3-1)(a2012-1)+(a2012-1) 2]t+2014t

Then there is g(t)=4028, and because g(2)=4028

a3-1)^2-(a3-1)(a2012-1)+(a2012-1)^2]＞0

So g(t) is an additive function with respect to t.

So t=2, i.e., a3+a2012=2

To sum up, choose A

The tolerance should be 3, sn=a1+(a2+an)*(n-1) 2

That is, 30=a1+15*(n-1) 2, from the meaning, n is greater than 1, and a1 is less than an less than 15, so only a1=0, n=5 is true, and d=3 is solved

As you can see from the title: sn=a(a[n]-1) (a-1) (where [n] represents the subscript of the series).

So s[n-1]=a(a[n-1]-1) (a-1)=a(a[n]-1) (a-1)-a[n] => a[n]=a(a[n]-a[n-1]) (a-1).

Simplified, it is obtained: a[n] a[n-1]=a So the number column is proportional to the series, and the common ratio is a

a[1]=s1 (a[1]-1) s1=(a-1) a => a[1]=a

So: a[n]=a n

an=a×（an-1），an/an-1=a

This is a proportional sequence that finds a1=a

According to the equation of the proportional series, an=a n can be found

You have a problem with this question, let's say n=1then a1+a2=1 2, contradicting s2=1.

s21=a1+a2+……a21

Because a2=1 so a1=-1 2

And a2 is added all the way to a21, and using the formula that the sequence of numbers told above satisfies it, we can see that this is the sum of 9 1 2, that is, subtract 1 2 to be the desired result, choose a

The sum of the two adjacent terms of the sequence is equal to 1 choice a

When an is selected as an odd number, an=—1 2

When n is an even number, an=1

an+a(n+1)=1 2,a(n+1)+a(n+2)=1 2, subtract from two formulas, and get an=a(n+2)Since a2=1, a1=-1 2, then a1=a3=a5....=a21=-1/2,a2=a4=a6=...

a20=1, so s21=11*a1+10*a2=9 2

From the meaning of the title, it can be seen that the sum of two consecutive terms is 1 2, and the second term is known to be 1, so the first term is -1 2

It can be obtained: when n is an odd number, an is -1 2

When n is even, an is 1

s21 = 11x(-1 2)+10x1=9 2, so choose item a.

When solving a number series problem, you should pay attention to the relationship between the items, and there are some skills such as sn+1 - sn = an, and pay attention to memorization.