What is the general formula of high school mathematical sequence 1, 1, 2,4,8,16,32? Solution:

When n= 1, a 1= 1.

When n≥2:

Starting from the second term, it is 1, 2, 4, 8, 16, 32, ... that is, a geometric series with a common ratio of 2.

Because the general formula is an = 2 (n-2)

Therefore, the general formula of this series is:

1 n= 1

Ann =

2^(n-2) n≥2

In senior high school mathematics, the general formula a 1 is the first term, an is the nth term, and D is the tolerance.

The first n terms and formulas are: sn = na1+n (n-1) d/2sn = (a1+an) n/2.

If m+n=p+q, then: am+an=ap+aq exists.

If m+n=2p, then: am+an=2ap.

All the above n.m.p.q are positive integers.

The general formula of (1) geometric series is: an = a 1× q (n- 1).

If the general formula is transformed into an = a 1/q * q n (n ∈ n *), when q > 0, an can be regarded as a function of independent variable n, and the point (n, an) is a set of isolated points on the curve y = a1/q * q X. 。

(2) the relationship between any two terms am and an an = am q (n-m).

(3) A1an = a2an-1= a3an-2 = … = akan-k+1,k ∈ {1 can be deduced from the definition of geometric series, the general term formula, the first n terms and the formula.

(4) Equal ratio mean: AQAP = Ar 2, Ar is AP, and AQ is equal ratio mean.

(5) Equal ratio summation: Sn=a 1+a2+a3+. +Ann

① when q≠ 1, sn = a1(1-q n)/(1-q) or sn = (a1-an× q) ÷ (/kloc)

② when q= 1, Sn=n×a 1(q= 1).

If π n = A 1 A2 … an, then π 2n- 1 = (an) 2n- 1, π 2n+1= (an+1) 2n+1.

In addition, each term is a geometric series with positive numbers, and the same base number constitutes an arithmetic number.

1/(2n- 1)

General formula of arithmetic progression: a(n)=a( 1)+(n- 1)d, where d is the tolerance;

General formula of geometric series: a (n) = a (1) * q (n- 1), where q is the common ratio;

The general formula an-1= a (n-1)+3 (n-1) in high school mathematics.

A(n- 1)- 1 = A(n-2)+3(n-2)

……=…………

A2- 1=A 1+3

an = n- 1+a 1+3[n(n- 1)/2]

=(3n^2-n)/2(n≥2)

In addition, A 1= 1 closes the formula.

So an = (3n 2-n)/2.

Because Sn+ 1=4an+2, the formula for finding the general term in high school mathematics

Sn=4an- 1 +2

Therefore an+1= 4an-4an-1an+1-2an = 2 (an-an-1) makes an+ 1-2an=bn+ 1.

So bn+ 1=2bn, so bn = b2 * 2 (n-2) = (a2-2a1) * 2 (n-2) = 3 * 2 (n-1).

So an-2an- 1 = 3 * 2 (n- 1)

an/2^n-an- 1/2^(n- 1)=3/2

So cn = an/2 n, so Cn-Cn- 1=3/2.

Cumulative cn = 3/2 (n-1) an = 3 (n-1) 2 (n-1) (n ≥ 2) a1=1.

An+ 1=2An+ 1 find the general term formula. For a given equation, let n= 1. Because a 1= A 1 is easy to get: a2 = 2a1+1(1). Because an+ 1=2An+ 1, subtract an = 2an-1:an+1-an = 2 (an-an-1). =2an, that is, an+ 1 = 3an. So the sequence an is the order n= 1 the first term is a 1 common ratio 3: a2 = 3a 1 and (1) are solved simultaneously: a 1 = 1+.

In high school mathematics, find the solution of the general term formula of sequence;

The characteristic equation is:

X^2=X+ 1

solve

X 1=( 1+√5)/2，，X2=( 1-√5)/2 .

Then a (n) = c1* x1n+C2 * x2n.

∫a( 1)= a(2)= 1 .

∴c 1*x 1+C2 * x2 = 1 .

c 1*x 1^2+c2*x2^2= 1。

The solution is C 1=√5/5, C2=-√5/5.

∴ A (n) = (√ 5/5) * {[(1+√ 5)/2] n-[(1-√ 5)/2] n} (√ 5 stands for the radical number 5).

High school mathematics: solving the general formula of series "First, switch n and An- 1 to An/An- 1=n/n- 1 and then write an-1/an-2 = n-1/n-2 until.