。。Urgent math accumulation, what is accumulation?

<> is not an analytic formula for an+1.

After listing the equation, the left and right sides can eliminate a large piece, written incorrectly on the paper, I haven't done high school math for a long time, followed by an equal proportional number series, plus a constant number series, an+1=2+(1-3 n) 1-3+2(n-1)=3 n 2+2n-1 2

3 n is to the nth power of 3.

an+1=an +3n + 2

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

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

a2 =a1 +3 + 2

The equation is added left and right.

an+1=a1+ 3(n+（n-1)+.1) +2n=3n(n+1)/2+2(n+1)

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

Accumulation: an-an-1=3(n-1)+2

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

an-2-an-3=3(n-3)+2

Arrive. a2-a1=3*1+2=5

So all add to :

an-a1=5+……3(n-3)+2+3(n-2)+2+3(n-1)+2=2(n-1)+(n-1)(3+3n-3)/2=3n(n-1)/2+2(n-1)

So an=a1+3n(n-1) 2+2(n-1)=2n+3n(n-1) 2

So an=2n+3n(n-1) 2=(3n 2+n) 2 The answer is accumulation, an 1 an=f(n) can have an (an an 1) (an 1 an 2) when a certain law is satisfied....a2 a1) This number is all added up and simplified. When writing the {an} general formula, it is important to verify that n=1 satisfies the case of an(n 2). If it is not satisfied, it is represented in segments.

From an+1=an+3n+2, an+1-an=3n+2, we have an - an-1=3(n-1)+2an-1 - an-2=3(n-2)+2 .........a2-a1=3*1+2 is added to obtain an-a1=3(n-1)+2+3(n-2)+2+......3*1+2=3*[1+2+3+……n-1)] +2*(n-1)=[3*(n-1)*n] 2+2(n-1)so:an=(3n2+n) 2

an+1=an+3n+2∴an+1-an=3n+2∴a2-a1=3+2a3-a2=6+2a4-a3=9+2………an-an-1=3(n-1)+2 is added to obtain a2-a1+a3-a2+a4-a3+......an-an-1=3+9+……3(n-1)+2(n-1)∴an-a1=(3+3n-3)x(n-1)/2+2(n-1)∴an=(3n/2+2)(n-1)+2，n∈n*

The accumulation method is also known as "chain eggplant."EquationsThe method of "algebraic averaging" refers to the use of an equation to express the consistency between the cumulative sum of the levels of each period calculated at the average rate of development from the initial level and the sum of the actual levels of the corresponding periods.

The equation method is used to calculate the average rate of development, focusing on the sum of the levels of each period of the medium- and long-term plan, that is, the cumulative total of the plan period. This method is suitable for calculating the amount of investment in the construction of the foundation and the addition of fixed assets.

amount, residential construction area, afforestation area, etc.; The average rate of development of the indicator.

Its basic starting point is: from time series.

The initial development level of A0 begins, with the average speed of the series.

In place of the month-on-month development rate of each period, the sum of the theoretical development levels of each period is consistent with the sum of the actual development levels of each period, that is:

To solve this higher-order equation, the positive root is the average rate of development. However, it is very troublesome to solve this higher-order equation, so in practical work, Lipeng Gaocha often uses the "Average Growth Rate Checklist" that has been compiled to calculate.

It can be seen that the equation method is used to calculate the average development rate, focusing on the sum of the levels of each period of the medium- and long-term plan, that is, the cumulative total amount during the plan period. This method is suitable for calculating the amount of capital investment, the amount of new fixed assets, the construction area of residential buildings, the area of afforestation, etc.; The average rate of development of the indicator.

1. Method:

2+100）x50/2

2x50 2+100x50 Collapse Bright Circle 2

2. Accumulation formula for addition:

n+(n+k)+(n+2k)+(n+3k)+、n+mk)[n+(n+mk)]*m/2

3. Principle: The first item plus the last item of the key beat is taken as the average, and then the collapse is multiplied by the number, such as , a total of 9 numbers, the first and the last one are added to equal to 10, and the average number is taken. Multiply by the average number of 9

Equal to 45(This seems to be junior high school math).

For example, the general term is an= 1 n - 1 (n+1) and find sn !

This is where the accumulation method comes in

a1=1 - 1/2

a2=1/2 - 1/3

a3=1/3 - 1/4

a4=1/4 - 1/5

a(n-1)=1/(n-1) -1/n

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

You can see it...sn= a1+a2+a3+..a(n-1)+an

It is equal to = 1 - (1 2) + (1 2) - (1 3) + (1 3) ...1/n)+(1/n)-[1/(n+1)]

Good covenant class .As a result, all that is left is 1- [1 (n+1)] !So that's the use of the accumulation method!

The specific branch rush of the accumulation method is a fierce skin.

For example, the general term is an= 1 n - 1 (n+1) and find sn !

This is where the accumulation method comes in

a1=1 - 1/2

a2=1/2 - 1/3

a3=1/3 - 1/4

a4=1/4 - 1/5

a(n-1)=1/(n-1) -1/n

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

You can see it...sn= a1+a2+a3+..a(n-1)+an

It is equal to = 1 - (1 2) + (1 2) - (1 3) + (1 3) ...1/n)+(1/n)-[1/(n+1)]

Good covenant class .As a result, all that is left is 1- [1 (n+1)] !So that's the use of the accumulation method!

What are the specific steps of the accumulation method.

What are the specific steps of the accumulation method.