An example of dislocation subtraction in high school mathematics sequence

Dislocation Subtraction Dislocation Subtraction is a common summation method of sequence, which is applied to the multiplication of geometric progression and arithmetic progression.

As An=BnCn, where Bn is arithmetic progression and Cn is geometric progression; List Sn separately, and then multiply all formulas by the common ratio of geometric series at the same time, that is, kSn；; Then an error occurs, and the two are subtracted.

For example, sum sn = x+3x+5x2+7x3+…+(2n-1) * x (n-1) (x ≠ 0).

When x= 1, sn =1+3+5+…+(2n-1) = n 2;

When x is not equal to 1, sn =1+3x+5x2+7x3+…+(2n-1) * x (n-1);

∴xsn=x+3x^2+5x^3+7x^4+…+(2n- 1)*x^n；

Subtract the two expressions to get (1-x) sn =1+2x [1+x+x2+x3+…+x (n-2)]-(2n-1) * xn;

Simplified to Sn = (2n-1) * x (n+1)-(2n+1) * x n+(1+x)/(1-x) 2.

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

Multiply both sides by 1/2.

1/2sn =1/4+1/8 ...+1/2n+1/2 (n+1) (note the difference from the original position, so write it.

Subtract two expressions

1/2sn= 1/2- 1/2^(n+ 1)

Sn= 1- 1/2^n

Dislocation subtraction is the solution of summation. In the question type: generally, it can only be used when the coefficient before a is equal to the index of a. This is an example (format problem, the numbers after a and n are exponents):

s = a+2 a2+3 a3+……+(n-2)an-2+(n- 1)an- 1+nan( 1)

Multiply both sides of (1) by a. Equation (2) is obtained as follows:

aS = a2+2 a3+3 a4+……+(n-2)an- 1+(n- 1)an+nan+ 1(2)

Using (1)-(2), Equation (3) is obtained as follows:

( 1-a)S = a+(2- 1)a2+(3-2)a3+……+(n-n+ 1)an-nan+ 1(3)

( 1-a)S = a+a2+a3+……+an- 1+an-nan+ 1

S = A+A2+A3+...+An- 1+An Use this summation formula.

( 1-a)S = a+a2+a3+……+an- 1+an-nan+ 1

Finally, divide by (1-a) on both sides of the equation at the same time, and you can get the general formula of S.

Example: sum Sn=3x+5x square +7x cube+ .........................................................................................................................................................

Solution: When x= 1, Sn =1+3+5+...+(2n-1) = the square of n.

When x is not equal to 1, Sn=Sn=3x+5x square +7x cube+...+(2n- 1) times x's n-1power.

So xSn=x+3x square +5x cube +7x four times ... +(2n- 1) times x n power.

So (1-x) sn =1+2x (1+the square of x+the cube of x+. . . . . +x to the power of n-2) -(2n- 1) times x to the power of n.

To put it simply: Sn=(2n- 1) multiplied by x is the square of n+ 1 power -(2n+ 1) multiplied by x's n power+(1+x)/(kloc-0/-x).

Cn=(2n+ 1)*2^n

Sn=3*2+5*4+7*8+...+(2n+ 1)*2^n

2Sn= 3*4+5*8+7* 16+...+(2n- 1)*2^n+(2n+ 1)*2^(n+ 1)

Subtract these two expressions.

-Sn=6+2*4+2*8+2* 16+...+2*2^n-(2n+ 1)*2^(n+ 1)

=6+2*(4+8+ 16+...+2^n)-(2n+ 1)*2^(n+ 1)

= 6+2 (n+2)-8-(2n+1) * 2 (n+1) (geometric series sum)