High School Mathematics Known sequences in a1 1, an a n 1 3 n 1 n 2 and n is a positive integer .

It's not that there is a formula, I haven't seen it for a long time, I don't remember, it's easy to solve a problem like this with a formula.

I haven't done it for two years, and I forgot about it! Just apply the formula! This type of question is often tested, and if you know how to do this kind of question, you won't worry about it in the future!

Your topic doesn't seem to be complete.

an=p, p is an odd number, so a(n+1)=5p+27 is an even number, and the good object is a(n+2)=(5p+27) 2 k p, so it is arranged into p*(2 k-5)=27, so 2 k-5=1,3,9,27, and because k is a positive integer sock, so 2 k-5=3,27So the attack liquid p = 1 or 9

a(n+1)=an+n

a(n+1)-an=n

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

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

a2-a1=1

Accumulate an-a1 = 1 + 2 + .n-1)an=a1+1+2+..n-1)=1+n(n-1) 2n=1, a1=1, is also satisfied.

The general formula for the series is an=1+n(n-1) 2n=100.

a100=1+100×99/2=4951

a2=a1+1=1+1

a3=a2+2=1+1+2

a4=a3+3=1+1+2+3

So a100 = 1 + 1 + 2 + 3....99=4951

Hello, if you don't understand, just ask, hehe.

Khan was wrong sorry for the change.

This is a proportional sequence with a common ratio of q to 2, and the formula of the proportional sequence can be used to cover the Hu an=a1*q (n 1) and sn=a1(1-q n) macros to destroy (1-q) to find Yu You.

s10=1(1-2^10)/(1-2)=2^10-1=1023

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

A(n-1)-a(n-2)=1-2(n-1)a(n-2)-a(n-3)=1-2(n-2)a3-a2=1-2 3

a2-a1=1-2×2

The above is added up.

an-a1=[1-2n]+[1-2(n-1)]+1-2(n-2)]+1-2×3]+[1-2×2]

n-1)-2[n+(n-1)+(n-2)…And friends....+3+2] in curly braces to a family of equal differences.

an-a1=(n-1)-2[(n²+n-2)/2]=n-1-(n²+n-2)=n-1-n²-n+2=1-n²

a1=1an=2-n²

a1=1

a2=a1+1

a3=a2+2

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

Add the above equations together;

a1+a2+……an=a1+a2+……a(n-1)+1+1+2+……n-1

sn=s(n-1)+1+2+……n

sn-s(n-1)=1+2+3+……n=n(n+1) 2Because sn-s(n-1)=an, so;

an=n(n+1)/2

a100=100(100+1)/2

Solution: (1) is composed of a(n+1)=an+2 n+1 a(n+1) -2 (n+1)=an - 2 n+1[a(n+1) -2 (n+1)] an - 2 n) = 1

Therefore, {an-2 n} can be regarded as a series of equal differences where the first term is a1-2=0 and the tolerance is d=1.

2）an-2^n = （a1-2^1）+ n-1)d=n-1

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

an+1-2^n+1=an-2^n+1

Shift the term to get (an+1-2 n+1)-(an-2 n)=1 so it is an equal difference series with a tolerance of 1

2) Let bn=an-2 n

then the series of equal difference from bn gives b1=a1-2=0bn=n-1=an-2 n

an=2 n+n-1

After testing, the above equation also holds when n=1, so an=2 n+n-1

1.Proof: Let b n = a n-2 n

then b (n+1)-b n=a (n+1)-2 (n+1)-a n+2 n

a n+2 n+1-2 (n+1)-a n+2 n=1, so the sequence {b n} i.e. {a n-2 n} is the difference series.

2.As we know from the above question, the number series {a n-2 n} is an equal difference series, the tolerance is d=1, and the first term is a 1-2 1=0

So a n-2 n=n-1, i.e. a n=2 n+n-1

an = n(n+1)/2

Mathematical induction is sufficient.

an = an/a(n-1) *a(n-1)/a(n-2) *a3/a2 * a2/a1 * a1

n+1)/(n-1) *n/(n-2) *n-1)/(n-3) *4/2 * 3/1 * 1

n+1)*n2 (the middle ones are gone).

a(n+1)=an/（2an+1）

Take the bottom of both sides.

1 a(n+1)=(2an+1) an=2+1 an, i.e., 1 a(n+1)=2+1 an

So the number column is an equal difference series with a tolerance of 2

So 1 an=1 a1+2(n-1)=2n-1an=1 (2n-1).

Because a1=1, an is not equal to 0;

a(n+1)=an/(2an+1)

1 a(n+1)=1 an+2 (the above equation is taken as the reciprocal), that is, 1 an=1 a(n-1)+2, the number column is an equal difference series with 1 as the first term and a tolerance of 2.