How to find the sum of proportional series, and the sum of equal difference series

Equal difference series sn=na1+n(n-1)d 2 or sn=n(a1+an) 2. The sum formula for the first n terms of the proportional series is: sn=[a1(1-q n)] (1-q) and the relation between any two terms am, an is an=am·q (n-m).

multiply by the common ratio) and then use the dislocation subtraction method.

The shape is an=bncn, where it is an equal difference series, which is an equal proportional series; List sn separately, and then multiply all the formulas by the common ratio q of the equal closure ratio sequence, i.e., q·sn; Then stagger one digit and subtract the two formulas. This method of summing the sequence is called dislocation subtraction.

Example]: Summation sn=1+3x+5x2+7x3+....+2n-1)·xn-1（x≠0，n∈n*）

When x=1, sn=1+3+5+....+2n-1)=n2When x≠1, sn=1+3x+5x2+7x3+....+2n-1)xn-1xsn=x+3x2+5x3+7x4+…+2n-1)xn subtracted to give (1-x)sn=1+2(x+x2+x3+x4+....+xn-1)-(2n-1)xn

Proportional Sequence Formula:

1. Definition:

2. Summation formula:

<>3. General term formula:

4. From the definition of proportional series, general term formulas, the first n terms and formulas, it can be deduced:

Equation for the Difference Series:

1. Definition:

For the series if satisfied:

The column is called an equal difference series. where the tolerance d is a constant and n is a positive integer.

2. General formula.

an=a1+(n-1)*d。The first term a1 = 1, tolerance d = 2.

3. The formula for the first n terms is: sn=a1*n+[n*(n-1)*d] 2sn=[n*(a1+an)] 2

sn=d/2*n²+（a1-d/2）*n

The method of summing the parallel items is often used to try first and then sum.

For example: 1 2+3 4+5 6+......Method 1: (merger)

Find the sum of odd and even terms, and subtract them.

Method 2: 1 2) + (3 4) + (5 6) + ....2n-1)-2n] Method 3: Construct a new series, which can borrow the compound of the equal difference series and the proportional series.

an=n(-1)^（n+1）

Extended information: 1. Formula summation method:

The formula for summing the equal difference series and the equal Li early ratio series.

Important formula: 1+2+....+n=

n（n+1）；

nn(orange n+1)(2n+1);

n（1+2+…+n）

nn+1)2, the method of summation of split terms: the general term of the sequence is divided into the algebraic sum of two formulas, that is, anf(n+1)-f(n), and then the sum of many terms in the middle is eliminated

anb)(anc)

c-banban+c

n(n+1)

nn+13, dislocation subtraction method: for the sum of the first n terms of a series composed of the product of the corresponding terms of the equal difference series and the proportional series, the dislocation subtraction method anb is commonly used

ncn where {b

n is the series of equal differences, {c. }

n is a proportional series.

4. Reverse order addition: s

n denotes the sum of the first term to the nth term, and then the sn is expressed as the sum of the nth term in reverse order to the first term, and the resulting two formulas are added together to obtain a summation method of sn.

The formula for the sum of odd terms in the equal difference series is: s odd = (a+nd)(n+1) The formula for the sum of even terms in the equal difference series is: s even = (a+nd)n The summing process is:

Let the first term of the original series be a, the tolerance be d, and the number of terms be 2n+1.

Then the original number series of hands and spikes are in order: A, A+D, A+2D, and the clan stares A+3D ......a+2nd

The odd terms are: a, a+2d, a+4d, ......a+2nd is calculated according to the formula of the equal difference series: sn=(the first term is called + the last term) * the number of terms 2 odd terms and the sum is:

Odd = A + A + 2nd)] (n+1) 2 = a+nd)(n+1)

The even terms are: A+D, A+3D, A+5D, ......a+(2n-1)d even terms and: s even = a+d) +a+2nd-d)]n 2 = a+nd)n

s odd s even = n+1) n

The common referent of the sum of the difference series is: sn= n* a1 + n*(n-1)d2 = n( a1+ an) 2

The formula for finding the sum of the equal Tan ratio series: sn= a1*( 1- q n) (1-q) =a1 - an*q) (1-q).

The method of summing the equal ratio sequence is as follows: Yuzi vertical

If the first term of the series is a1, the tolerance is d, the nth term is an, and the sum of the first n terms is sn, then there is: sn = n 2 [2a1 + n-1)d].

According to the meaning of **, it should be the sum of the first n terms of the sequence {cn}, where cn=(2 (n-1)) (2n-1), which is the proportional series divided by an equal difference series, and the sum of the first n terms cannot be found. Only by multiplying the equal difference series by the proportional series can the sum of the first n terms be found by dislocation subtraction. Try to sum, such as **, but the final result cannot be simplified.

Whose sum is it for? The question seems incomplete.