Known sequence 1,2,3,4 2n, find the sum of them

Known sequence 1,2,3,4 2n, find the sum of them

1,+2+3+4+…… +2n
=(1+2n)+(2+2n-1)+(3+2n-2)+…… +(n+2n-n)
=(1+2n)(n-1)+n
=n-1+2n^2-2n+n
=2n^2 -1

Given the tolerance of {an} sequence D ＞ 0, A4 + A6 = 10, A4 × A6 = 24. Find {an general term formula}
Within half an hour, which great Xia made an additional 10 points! I'm in a hurry, online

a4+a6＝10...（1）
a4×a6＝24...（2）
The results show that A4 = 4, A6 = 6
d=（a6-a4）/2=(6-4)/2=1
a4=a1+3d
a1=a4-3d=4-3*1=1
The formula is an = a1 + 1 * (n-1) = 1 + n-1 = n

It is known that the tolerance D of arithmetic sequence an is greater than 0, and A4 + A6 = 10, A4 * A6 = 24
BN = 1 / an * a n + 1, the sum of the first n terms of BN is TN, TN ≥ m, find the maximum of integer M

Tolerance d > 0, and A4 + A6 = 10, A4 * A6 = 24, so A4 = 4, A6 = 6, tolerance = (a6-a4) / 2 = 1,
a4=a1+3d
a1=1
an=1+(n-1)d=n
bn=1/(n^2)+1
Tn=(1/1^2+1)+(1/2^2+1)+...+(1/n^2+1)=(1/1^2+1/2^2+...1/n^2)+n
≥[1/(1*2)+1/(2*3)+...+1/(n*(n+1))]+n
=[(1-1/2)+(1/2-1/3)+...+(1/n-1/(n+1))]+n
=1-1/(n+1)+n
So the maximum of M is n + 1

Given that the tolerance of the arithmetic sequence an is a positive number, and a3a7 = - 12, A4 + A6 = - 4, then S20=
I want ideas. Ideas for doing questions

A4 + A6 = A3 + A7, then according to the first condition, we can find A6 = 2, A3 = - 6 or A3 = 2, A6 = - 6 (according to the tolerance is a positive number, rounding off), then we can find A1 and tolerance, and then we can find S20 according to the formula

Given the first n terms of arithmetic sequence {an} and Sn = - 2n ^ 2-n
a. Find the expression of the general term an;
b. Find a1 + a3 + A5 +The value of a25
Please answer this question in detail and write down the specific part

Sn = - 2n ^ 2-ns (n-1) = - 2 (n-1) ^ 2 - (n-1) an = SN-S (n-1) = - 2n ^ 2-N + 2 (n-1) ^ 2 + (n-1) = 2 [(n-1) ^ 2-N ^ 2] - 1 = - 4N + 1A1 = - 3an is an arithmetic sequence with - 3 as the first term and - 4 as the tolerance, then A1, A3 It is an arithmetic sequence with - 3 as the first term and - 8 as the tolerance, with a total of (25-1) / 2 + 1 = 13 terms, so a1 + a3 +

Given the first n terms of sequence an and Sn = 2n2-5n + A + 1, if an is an arithmetic sequence, then a is a real number=

a+1=0,a=-1

If {Sn / an} is an arithmetic sequence with tolerance D, then {an} is an arithmetic sequence if and only if d =?
From Sn / an = 1 + (n-1) d (the first term is S1 / A1 = 1), it is obtained that: 1
S2 / A2 = 1 + D, A2 = A1 / D, A3 = (a1 + A1 / D) / 2D,
From 2A2 = a1 + A3 (A1 is not equal to 0), we can get: 2D ^ 2-3D + 1 = 0, d = 1 or D = 1 / 2
Why is this solution tested? Why is it a special value?

Because D obtained in this way can only guarantee 2A2 = a1 + a3, that is to say, the first three terms form an arithmetic sequence, but not the last three terms
It can be analyzed in a more general way

It is known that the sum of the first n terms of the sequence {an} is Sn = (n + 1) 2 + T. it is proved that the necessary and sufficient condition for {an} to be an arithmetic sequence is t = - 1

Sn=(n+1)^2+t,a1=S1=4+t；
So when n > 1, an = SN-S (n-1) = 2n + 1
If the sequence is an arithmetic sequence, A1 also conforms to an = 2n + 1
So 4 + T = 3, t = - 1,
That is to say, t = - 1 is a necessary condition for (an) to be an arithmetic sequence

Given the first n terms of sequence {an} and Sn = an2 + BN + C (a ≠ 0), the necessary and sufficient conditions for sequence {an} to be an arithmetic sequence are obtained

When n = 1, A1 = a + B + C; when n ≥ 2, an = sn-sn-1 = 2An + b-a. because a ≠ 0, when n ≥ 2, {an} is an arithmetic sequence with tolerance of 2A. If {an} is an arithmetic sequence, then a2-a1 = 2A, the solution is C = 0. The necessary condition for {an} to be an arithmetic sequence is a ≠ 0, C = 0. Sufficiency: when a ≠ 0, C = 0, Sn = an2 + BN. When n = 1, A1 = a + B; when n ≥ 2, an = sn-sn-1 = 2An + B-A, obviously when n = 1 In conclusion, it can be seen that the sequence {an} is an arithmetic sequence if and only if a ≠ 0, C = 0

It is proved that the sequence {an} is an arithmetic sequence if and only if its first n terms and Sn = an ^ 2 + BN (constant a, B ∈ R) are equal

Sufficiency: an = SN-S (n-1) = a (2n-1) + B = 2A n + B-A
d=an-a(n-1)=2a
necessity:
Let the first term of the arithmetic sequence be A1 and the tolerance be d,
Then:
Sn=a1n+n(n-1)d/2=(d/2)n^2+n(a1-d/2)
a=d/2,b=a1-d/2