Solve a super difficult number series puzzle

a1=2a2=2*2+2^1=6

a3=2*6+2^2=16

a4=2*16+2^3=40

a5=2*40+2^4=96

By observation, we can assume that an=(n+1)*2 (n-1) (n>=2).

Let's start the argument and assume that an=(n+1)*2 (n-1) is true for the natural number n>=2.

an+12an+2^n

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

n+1)*2^n+2^n

n+2)*2^n

is consistent with the above assumptions, so the assumptions are valid.

Then the general formula is.

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

The bottom part is the process!! Then I'll delete the idea of doing the question.

Superposition method: a(n+1)=2a(n)+2 n

2a(n)=4a(n-1)+2^n

4a(n)=8a(n-1)+2^n

2 (n-2)a(3)=2 (n-1) a2+2 n2 (n-1)a(2)=2 n a1+2 n The above formulas are added to remove the same terms on the left and right.

a(n+1)=2^n×a1+n×2^n

Because a1=2

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

i.e. an=(n+1)2 (n-1).

Since a(n+1)=2a(n)+2 n is divided by 2 (n+1) on both sides of the equation, a(n+1) 2 (n+1)=2a(n) 2 (n+1)+2 n 2 (n+1)=a(n) 2 n+1 2Let b(n)=a(n) 2 n be obtained from the previous equation b(n+1)=b(n)+1 2 and b(1)=a(1) 2 1=2 2=1So b(n) is a series of equal differences with 1 as the first term and 1 2 as the tolerance, so b(n)=1+(n-1)(1 2)=(n+1) 2=a(n) 2 n, so a(n)=(2 n)(n+1) 2=(n+1)2 (n-1).

In junior high school competitions, this kind of question is listed first, and then observed, and the general rules are found to be written.

Again, this is the general process.

The second floor is considered proof, but in the end it should be added that it is based on mathematical induction.

The 3rd floor can only be regarded as thinking and reasoning, and it is estimated that it will be difficult to score if you really answer the question.

I'll do it anyway. I saw you faint to death.

a1=2, a2=1 2, a3=-1, a4=2, and then recycle, the period t=3 is a(n+3)=an, according to the general formula.

sin(wn+3w+)=sin(wn+)3w=2, w=2 3, by the general formula n respectively in the case of .

1、asin(2π/3+φ)b=2

2、asin(4π/3+φ)b=1/2

3. ASIN(2 + b=-1,-- is ASIN +b=2

The formalization is simplified as:

a(-1 2 sin + 3 2 cos) + b=2a(-1 2 sin - 3 2 cos) + b=1 2 The addition of the two formulas gives -asin +2b=5 2

Then it is obtained by simultaneous equation with equation 3.

b = 1 2, asin = -3 2, bring them into the formula 1 or 2, and find them out at lightning speed.

acosφ=√3/2

Add ASIN = -3 2 and ACOS = 3 2 squares to quickly find A= 3, and then quickly find out the bell.

sinφ=-√3/2

By your conditions, |Half of the pies.

At this point, we have the full expression.

an= 3sin(2n 3- 3)+1 2= 3sin[(2n-1) 3]+1 2You can check the correctness of the first three terms on behalf of n.

Cow b.,As long as you're careful and don't make mistakes.,Rejoicing.,It's so simple I can't help but add.,I feel like I've solved it too wonderfully.。。

an=1-1 a(n-1).

an=asin(wn+ )b is a general term.

What do you want?

Standard Answer:

an=sn-sn-1(n>=2)

an=1 2a(n-1)-1 2a(n-2)=(1 2)a substituting a=1 into an does not match, then the sequence is segmented to form an=1 (when n=1), an=1 2a(n>=2).

a(n-1) should be the (n-1) item, it feels difficult, to use the competition method of high school mathematics to do, the name is the eigenroot equation method, for this problem is 2*a(n+2)=a(n+10)-an.You can search for this method on the Internet yourself, learn it, and it should be easy to make.

Does a(n-1) represent a multiplied by (n-1) or n-1 ah.

a(n+1)-an=2, which is a fixed value, and the number column is an equal difference series.

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

The first n term and sn=n*1+n(n-1)*2 2=n+n 2-n=n 2sn n=n 2 n=n

tn=1+2+..n=n(n+1)/21/tn=2/[n(n+1)]=2[1/n-1/(n+1)]sn=2[1-1/2+1/2-1/3+..1/(n-1)-1/n]

2(1-1/n)

2(n-1)/n

From a(n+1)=an+2, a(n+1)-an=2, so the series is a series of equal differences with 1 as the first term and a tolerance of 2.

And the sum of the first n terms is sn=na1+n(n-1)*d 2=n 2 Obviously, the series starts with 1 and has an equal difference with a tolerance of 1.

tn=n+n(n-1)/2=(n^2+n)/22.It can be derived from 1.

The general term of the series is 2 (n 2+n) = 2 n (n + 1) = 2 * 1 n (n + 1).

2*[1/n-1/(n+1)]

Then the sum of the first n terms of the series = 2*[1-1 2+1 2-1 3+1 3-1 4+.1/n-1/(n+1)]=2*[1-1/(n+1)]=2n/(n+1)