About the method of summation of superposition multiplication of proportional sequences.

an- an-1=n

an-1 - an-2=n-1

an-2 - an-3=n-2

a3-a2=3.

a2-a1=2

Adding the superstructure gives an -a1=2+3+4+.n so an=1+2+.n.

an-a(n-1)=n, then a(n-1)-a(n-2)=n-1, and so on a2-a1=2

Then the addition of the left side is [an-a(n-1)]+a(n-1)-a(n-2)].a2-a1]+a1=n+(n-1)..2+1

So an=1+2+3....n

In the same way, if the proportional sequence is an a(n-1)=q, q is the ratio, a(n-1) a(n-2)=q, then the multiplication on the left is extremely an=a1xq (n-1).

Take your example.

an-an-1=n

an-1-an-2=n-1

And so on. a3-a2=3

a2-a1=2

Adding the left and right gives an-an-1+an-1-an-2+......a3-a2+a2-a1=n+n-1+n-2+……3+2

Only an-a1 is left on the left, and it can be calculated by summing the equation of the equal difference sequence on the right.

(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 formulas by the common ratio q of the proportional series at the same time, that is, 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-1∴xsn=x+3x2+5x3+7x4+…+2n-1)xn subtracted to give (1-x)sn=1+2(x+x2+x3+x4+....+xn-1)-(2n-1)xn

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

For example, let sn=1*2+2*2 2+3*2 3++n*2 n (1) then 2*sn= 1*2 2+2*2 3+3*2 4++n-1）*2^n+n*2^(n+1) （2）

Then (2)-(1) gets: 2*sn-sn=n*2 (n+1)-2 1-2 2-2 3--2^n

The equation at the left end is simplified.

Summing can be achieved by using the dislocation subtraction method.

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

For example, let sn=1*2+2*2 2+3*2 3++n*2^n （1）

Then 2*sn= 1*2 2+2*2 3+3*2 4++n-1）*2^n+n*2^(n+1) （2）

Then (2)-(1) gets: 2*sn-sn=n*2 (n+1)-2 1-2 2-2 3--2^n

The equation at the left end is simplified.

You asked: What is the formula for summing the product of the corresponding terms of the equal difference series and the proportional series?

The method is to multiply each term of the series that requires the sum to be the common ratio of the "proportional series", and then subtract the wrong terms to get a problem that can be summed by the "proportional series", and then simplify it!

1. Accumulation method (superimposed addition).

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

You can ask for Luchai and.

What is it? Arithmetic progression.

Mixed and complex points of proportional sequences plus 1 2+2 2+3 3+...n 2 or the sum of the cubes, within the range you know.

2. Accumulation method an a(n-1) = f(n).

Where f(n) does it need to be like the accumulation method (where f(n) is a series of equal differences or proportional numbers or other summable requirements?

In the case of the accumulation method, f(n) is generally clever and flexible, such as an a(n-1)=(n-1) (n+1), and there is basically no f(n) as an equal difference series or proportional series or other sums like the accumulation method.

Multiplication is actually a simple algorithm for the addition of the same number of additions, so it is relatively simple to find the sum of several identical additions with multiplication For example: 2+2+2=6

2*3=6A.

Multiplication is a simple algorithm for finding the sum of the same additions, so it can be calculated using multiplication.

Finding the sum of the same addition can be calculated by multiplication.

According to: the definition of multiplication - the simple operation of finding the sum of several identical additions is called multiplication.

Example: 2 2 2 2 2 2, 5 2's are added together, which is 2 times 5.

2+2+2+2+2=2x5=10。

Add several numbers to the same add.

Finding the sum of the same addition can be calculated by multiplication.

Exactly.