The sum of the series is a quadratic function, and whether it can be proved to be a series of equal

No. Let sn=a*n 2+b*n+c

then an=sn-s(n-1)=a*n 2+b*n+c-[a*(n-1) 2+b*(n-1)+c].

a*(2*n-1)+b

So a1=a*(2*1-1)+b=a+b (1) and a1=s1=a+b+c (2).

So make (1)=(2), c=0

Therefore, for a general quadratic function, it cannot be proved to be a series of equal differences.

Let the sum of the first n terms of the series an sn=xn 2+yn+z (is a function of n, x, y, z are constants, and x≠0).

s(n-1)=x(n-1)^2+y(n-1)+zan=sn-s(n-1)

xn^2+yn+z]-[x(n-1)^2+y(n-1)+z]xn^2-x(n-1)^2+y

2xn-x+y

2x(n-1)+y+x

x+y)+2x(n-1)

a1=s1=x+y+z

a2=3x+y

That is, the first term is (x+y+z), and when n 2, the number column is an equal difference series with a tolerance of 2x.

sn = a1*n+n*(n-1)*d/2 = n^2*d/2+(a1-d/2)*n

Therefore, only a quadratic function with a constant term of 0 can deduce a series of equal differences. And it's a if and only if relationship.

The sum of the first n terms of the equal difference series is sn=a1n+[n(n-1)d] 2=(d 2)n 2+(a1-d 2)n, where the quadratic term coefficient is .

d 2, the coefficient of the primary term is a1-d 2.

If there is a summation formula sn=an 2+bn+c, where a, b, and c are quadratic coefficients, primary coefficients, and constants, respectively.

When a=d 2, b=a1-d 2, c=0, this series is a series of equal differences.

Summary. The first n terms and formulas of the series of equal differences are about the quadratic function of n, and the constant term of this quadratic function is 0

The sum of the first n terms of the equivariance series is about the quadratic function of n, and the constant term of this quadratic function is 0sn=na1+d*n(n-1) 2=d 2*n 2+(a1-d 2)n, so the constant term of the quadratic function is 0

The sum of the first n terms of the equal difference series is sn=a1n+[n(n-1)d] 2=(d 2)n 2+(a1-d 2)n, where the quadratic term coefficient is .

d 2, the coefficient of the primary term is a1-d 2.

If there is a summation formula sn=an 2+bn+c, where a, b, and c are quadratic coefficients, primary coefficients, and constants, respectively.

When a=d 2, b=a1-d 2, c=0, this series is a series of equal differences.

- How to prove the series of equal delay and the number of the equal delay Allow the series of equal difference numbers an=a1+(n-1)d to divide the largest number plus the minimum number by two, i.e., [a1+a1+(n-1)d] 2=a1+(n-1)d 2 The mean of 2 is sn n=[na1+n(n-1)d 2] n=a1+(n-1)d 2 It is proved that 1 The three numbers abc are in an equal difference series, then c-b=b-a c 2(a+b)-b 2(c+a)=(c-b)(ac+bc+ab) b 2(c+c+a)-a2(b+c)=(b-a)(ac+bc+ab) Since c-b=b-a, then (c-b)(ac+bc+ab)=(b-a)(ac+bc+ab) i.e. c 2(a+b)-b 2(c+a)=b 2(c+a)-a 2(b+c) so a 2(b+c), b 2(c+a), c 2(a+b) into a series of equal differences Difference: an-(an-1)=constant (n 2) Proportional: an (an-1=constant (n 2) Equal Difference:

an-(an-1)=d or 2an=(an-1)+(an+1),(n2) proportional: an(an-1)=q or an-squared = (an-1)*(an+1)(n 2)2 We speculate that the general formula for a series of numbers is an=5n-4

The proof that this proposition is correct is as follows: Let the first n terms of the delay number and the talk be sn, and sn=na+n(n-1)d 2 where a and d are constants. When n=1, a1=s1=a is obtained from sn=na+n(n-1)d 2, and when n 2, an=sn-s(n-1)=[na+n(n-1)d 2]-[n-1)a+(n-1)(n-2)d 2]=a-d+dn, i.e., an=a-d+dn and a1=a....

For example, the difference between them is , and it is a series of equal differences.

This is called a quadratic difference series.

For example, 2, 4, 8, 14, 22 ......In this series, after making a difference between the two adjacent items, the new series 2, 4, 6, 8 ...... is obtainedThis is an equal difference series, and the second level of equal difference is to make a difference, and the new number column obtained is an equal difference series.

The general formula is: an

If an-a(n-1)=a, a is a constant number, it can be proved to be a series of equal differences.

For example, a(n+1)=8

an=6a(n-1)=4

a(n+1)-an=an-a(n-1)=2 in order to be provable.

The difference between two adjacent elements can form a new series, and if the new sequence is an equal difference series, then the original series is a quadratic equal difference series.

For example, 2, 5, 10, 17, 26....

The difference between the two adjacent items is 3, 5, 7, 11....is a series of equal differences, 2, 5, 10, 17, 26....It's called a quadratic difference series.

The sum of the first n terms of the equal difference series is sn=a1n+[n(n-1)d] 2=(d 2)n 2+(a1-d 2)n, where the quadratic term coefficient is .

d 2, the coefficient of the primary term is a1-d 2.

If there is a summation formula sn=an 2+bn+c, where a, b, and c are quadratic coefficients, primary coefficients, and constants, respectively.

When a=d 2, b=a1-d 2, c=0, this series is a series of equal differences.

No, to meet the definition of a series of equal differences, every two differences are fixed.

a[n]=a[1]+(n-1)d=dn+(a[1]-d)

The difference series is a line-like line composed of discrete equal x-axis spacing points with d as the slope and (a[1]-d) as the intercept.

Let sn=an 2+bn+c, then Ye Liangzai a1=s1=a+b+can=sn-s(n-1)=2an-(a-b)--an-a(n-1)=2a

This is a constant that states that from the second term onwards is a series of equal differences.

Whether the equation a1=a+b+c is 2a-(a-b) determines whether the first term is one of the equal difference series, obviously if c = 0, a1 is one of the equal difference series, otherwise, no. It can be seen that when c = 0, the number series is the equal difference series, and when c <> 0, the key number series is the equal difference series from the second term.