The first N terms of the proportional series and how to prove it

Answer: Let sn=a1+a2+.an

then qsn=a2+a3+.an+1

Subtract the two formulas, then (1-q)sn=a1-an+1=a1-a1*q n times.

We'll see you later!

However, pay attention to the discussion of cases where the common ratio is 1.

If the common ratio q=1, then sn=a1+a2+.ana1+a1+..a1=na1

The first n terms of the proportional series and sn=a1+a2+.ana1(1-q^n)/(1-q)

Ratio q≠1).

Certificate: SN=A1+A1Q+A1Q 2....a1q^(n-1)..1)qsn=a1q+a1q^2+..a1q^(n-1)+a1q^n...2)

1-q)sn=a1-a1q^n

sn=a1(1-q^n)/(1-q)

If the common ratio q=1, then sn=a1+a2+.ana1+a1+..a1=na1

The first n terms of the proportional series and sn=a1+a2+.ANA1(1-Q N) (1-Q)(Common Ratio Q≠1) Certificate: SN=A1+A1Q+A1Q 2....

a1q^(n-1)..1)qsn=a1q+a1q^2+..a1q^(n-1)+a1q^n...

2)(1)-(2):(1-q)sn=a1-a1q^n

sn=a1(1-q^n)/(1-q)

The relationship between q and 1 is quite important.

Geometric progression. The first n terms and formulas: sn = a1(1-q n) (1-q).

The derivation is as follows: because an = a1q (n-1) then baisn = a1+a1*q 1+.a1*q^(n-1) (1)

qsn =a1*q^1+a1q^2+..a1*q n (2)1)-(2) Note that the first term of Eq. (1) does not change.

Subtract the first term of Eq. (2) from the second term of Eq. (1).

Subtract the second term of Eq. (2) from the third term of Eq. (1).

By analogy, subtract the n-1 term of equation (1) from the n-1 term of equation (2).

2) The nth term of the equation remains unchanged, which is called dislocation subtraction, and its purpose is to eliminate this common term.

So I got it. 1-q)sn = a1(1-q n), i.e. sn = a1(1-q n) (1-q).

This is the summation of the common line of the proportional series, when the common ratio of the file fiber is q=1, sn=na1, when q≠1, sn=a1(1-q n) 1-q.

The first n terms of the proportional series and the formula are:

A proportional sequence formula is a mathematical formula for finding the sum of a certain number of proportional sequences. In addition, a proportional series where each item is positive is taken from the same base number to form an equal group of grandchildren. Conversely, if you take any positive number c as the base and use the terms of an equal difference series as an exponential power to construct can, it is an equal proportional series.

Proportional series properties

If m, n, p, q n*, and m+n=p+q, then am·an=ap·aq;

In a proportional series, when q≠-1, or q=-1 and k is an odd number, the sum of each k term is still a proportional series.

For example, banks have a way to pay interest --- compound interest.

That is, the interest of the previous period and the principal wheel hole are added together as the principal, and then the interest of the next period is calculated, which is what people often call the interest rolling.

The formula for calculating the sum of principal and interest according to compound interest: sum of principal and interest = principal (1 + interest rate) deposit period.

a1+..an=48,a(n+1)+.a2n=60-48=12, so the common ratio of the equal ratio is 1 4, then a(2n+1)+.

The first 3n terms of a3n=3 and =48+12+3=63 just thought it was a series of equal differences, but I read it wrong.

Yours is n-1 in total, not n.

The general formula is an=1 2 (n+1), so 1 2 n is obviously the n-1 term.

So it should be s(n-1) instead of sn

Neither is right!

Neither of them is the sum of the first n terms, but the sum of the first n-1 terms.

The first one, the second number or n, you have one less item...

1/4+1/8+..1/2^n

where q=1 2, 1 4=a1*q gives a1=1 2, then 1 2 n =a1*q (n-1), is the n-1 term of an.

s(n-1)=1/4+1/8+..1 2 n=1 2-1 2 n Summary rule: If the proportional series is proportional to 2 or 1 2, then the sum of the consecutive n terms of the series is: twice the value of the largest term minus the value of the smallest term.

sn=[a1*(1-q n)] 1-q) is an equal proportional series, and where n is an unknown number, it can be written as f(n)=[a1*(1-q n)] 1-q), when q=1 is a constant series, that is, n a1 is added to n*a1.

If the ratio of each term to the previous term from the second term of a series is equal to the same constant, the sequence is called a proportional sequence. This constant is called the common ratio of the proportional series, and the common ratio is usually represented by the letter q (q≠0).

Note: When q=1, an is a constant series. i.e. a n=a.

In general, if the ratio of each term to its preceding term from the second term of a series is equal to the same non-zero constant, the series is called an proportional series. This constant is called the common ratio of the equal-chaos Zheng Bi ratio sequence, and the common ratio is usually represented by the letter q (q≠0). Note:

When q=1, an is a constant series (n is a subscript).

[1＋a^(-1)

a^(-2)＋…a^(1-n)]

(3n-2)]

The former is a proportional series, and the common ratio is a (-1).

The latter is a series of equal differences with a tolerance of 3

1-a^(-n)]/1-a)

3n-2)]*n/2

1-a^(-n)]/1-a)

3n-1)n/2

Sum by the split term method.

This is the specific application of the idea of decomposition and combination in the summation of sequences.

The essence of the split term method is to decompose each term (general term) in the series and then recombine it so that some terms can be eliminated, and finally the purpose of summing is achieved.

General term decomposition (split term) such as:

1）1/n(n

1)=1/n-1/(n

2）1/(2n-1)(2n

1)=1/2[1/(2n-1)-1/(2n3）1/n(n

1)(n2)=1/2[1/n(n

1)-1/(n1)(n

4）1/(√a

b)=[1/(a-b)](a-√b)

n·n!=(n

1)!-n!

Example] Find the sequence an=1 n(n

The first n terms and.

Solution: Set. an=1/n(n

1)=1/n-1/(n

split). sn=1-1/2

1/n-1/(n

1) (Summing of split items) Bury He Carry.

1-1/(n

n (n summary: This type of deformation is characterized by the fact that after each term of the original sequence is split into two, most of the items in the middle of the pat cancel each other out. There are only a limited number of items left.

Note: The remaining items have the following characteristics.

1. The position of the remaining items is symmetrical.

2. The positive and negative properties before and after the remaining terms are reversed.

Answer: It can be obtained from the proportional vertical wisdom sequence.

a1=1,a4=1x(q) (4-1)=1 8 solution: q=1 2

So the first term is 1, and the common ratio is 1 2 in a proportional series, sn=(1-1 2 n) (1-1 2).

Therefore, if you bring in the sn formula, you can get sn=[1(1-1 2 10)] 1-1 2)=2-1 512=1023 512

sn=[a1*(1-q n)] (1-q) is a proportional series, and where n is an unknown number, it can be written as f(n)=[a1*(1-q n)] (1-q) when q=1 is a constant series, that is, n a1s are added together.

First, you use the sum of the first n terms of the proportional series, the premise is that q is not equal to -1, and then, if q = -1, it is a swinging series, that is, positive and negative alternately, such as 2, -2, 2, -2, 2, -2, ......sn=0, s2n=0, s3n=0, yes.

Solution: It can be obtained from the proportional series.

a1=1,a4=1x(q) (4-1)=1 8 solution: q=1 2

So the first term is 1, and the common ratio is 1 2 in a proportional series, sn=(1-1 2 n) (1-1 2).

Therefore, if you bring in the sn formula, you can get sn=[1(1-1 2 10)] 1-1 2)=2-1 512=1023 512

(1).a(n+1)=(1-%20)an+30%bn(2).bn=1000-an

then a(n+1)=(1-20%)an+30%(1000-an)=1 2*an+300

3).According to the formula proved in the second question, a(n+1)-600=1 2(an-600).

The result is a proportional series in the common ratio of 1 2, an-600 = (a1-600)*(1 2) (n-1).

an=(a-600)*(1/2)^(n-1)+600