Let the general term formula of sequence {an} be an = N2 + λ n (n ∈ n *) and {an} satisfy A1

Let the general term formula of sequence {an} be an = N2 + λ n (n ∈ n *) and {an} satisfy A1

The difference method is used
a(n+1)-a(n)
=(n+1)²+λ(n+1)-[n²+λn]
=2n+1+λ
Given the condition, {an} is an increasing sequence
The constant value of 2n + 1 + λ > 0 is tenable
The minimum value of ∵ 2n + 1 + λ is 2 * 1 + 1 + λ = 3 + λ > 0
∴ λ>-3
That is to say, the value range of real number λ is (- 3, + ∞)

The general term formula of known sequence {an} is an = N2 + kn + 2. If an + 1 > an holds for any n ∈ n *, then the value range of real number k is k > - 3
The general term formula of known sequence {an} is an = N2 + kn + 2. If an + 1 > an holds for any n ∈ n *, then the value range of real number k is ()
A．k>0 \x05\x05\x05B．k>－1
C．k>－2 \x05\x05 D．k>－3
From an + 1 > an, we know that the sequence is an increasing sequence, and because the general formula an = N2 + kn + 2 can be regarded as a quadratic function of N, considering n ∈ n *, so - k2-3
How do three-thirds of them come from?

Because an + 1 > an
So an + 1 - an = (n + 1) ^ 2 + (n + 1) K + 2-N ^ 2-kn-2 = 2n + 1 + k > 0
So k > - (2n + 1)
k>-3

Let the general term formula of the sequence {an} be an = n ^ 2-PN. If the sequence {an} is an increasing sequence, then the value range of the real number P is?

If an is an increasing sequence, then
A(n+1)-An=(n+1)²-p(n+1)-n²+pn=2n+1-p＞0
∴p＜2n+1
For any n ∈ n +, 2n + 1 is increasing
P is less than the minimum value of 2n + 1
When n = 1, the minimum value of 2n + 1 is 3
∴p＜3
This is what we need

The general formula of sequence {an} is an = n ^ 2-cn + 1. If an ≥ A3, then the value range of real number C is

Divided into two parts, consider A1, A2 ≥ A3 and an (n > 3) ≥ A3
The symmetry axis of the function is x = C / 2, and the limit case is C / 2 = 5 / 2 (A2 = A3) or C / 2 = 7 / 2 (A3 = A4) and C = 5 or 7
So 7 ≥ C ≥ 5

General term formula an = n & # 178;, find SN

Sn=1^2+2^2+3^2+...+(n-1)^2+n^2
Using the formula of cubic difference
n^3-(n-1)^3=1*[n^2+(n-1)^2+n(n-1)]
=n^2+(n-1)^2+n^2-n
=2*n^2+(n-1)^2-n
2^3-1^3=2*2^2+1^2-2
3^3-2^3=2*3^2+2^2-3
4^3-3^3=2*4^2+3^2-4
.
n^3-(n-1)^3=2*n^2+(n-1)^2-n
Add all the equations
n^3-1^3=2*(2^2+3^2+...+n^2)+[1^2+2^2+...+(n-1)^2]-(2+3+4+...+n)
n^3-1=2*(1^2+2^2+3^2+...+n^2)-2+[1^2+2^2+...+(n-1)^2+n^2]-n^2-(2+3+4+...+n)
n^3-1=3*(1^2+2^2+3^2+...+n^2)-2-n^2-(1+2+3+...+n)+1
n^3-1=3(1^2+2^2+...+n^2)-1-n^2-n(n+1)/2
3(1^2+2^2+...+n^2)=n^3+n^2+n(n+1)/2=(n/2)(2n^2+2n+n+1)
=(n/2)(n+1)(2n+1)
1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6
That is: SN = 1 ^ 2 + 2 ^ 2 + 3 ^ 2 +... + (n-1) ^ 2 + n ^ 2 = n (n + 1) (2n + 1) / 6

A1 = 1, Sn = n & # 178; × an, finding the general term formula of an

An = Sn - Sn-1 = n²×An - (n-1)²×An-1===> (n²-1)An = (n-1)²An-1===> (n+1)An = (n-1)An-1===> An = (n-1)/(n+1) An-1===> An = (n-1)/(n+1)×(n-2)/n ×(n-3)/(n-1)×…… × 3/5 × 2/4 ×...

The sum of the first n terms of an is Sn, 2Sn = an & # 178; + an, finding the general term formula of an

2A1 = 2S1 = A1 squared + A1, so A1 = 12sn = an squared + an2s (n-1) = a (n-1) squared + a (n-1) subtracted to obtain 2An = an squared + an-a (n-1) squared - A (n-1), that is, an squared - A (n-1) squared = an + a (n-1) so [an-a (n-1)] [an + a (n-1)] = an + a (n-1) so an-a (n-1) = 1

Given A1 = 1, Sn = (sn-1 + √ 2) &# 178;, find the general term formula of an
N-1 is below s,

Sn=(Sn-1+√2)²=（Sn-2+2√2)²=(Sn-3+3√2)²=...=[S1+（n-1）√2]²=[a1+（n-1）√2]²=[1+(n-1)√2]²∴Sn-1=[1+(n-2）√2]²an=Sn-Sn-1=[1+(n-1)√2]²-[1+(n-2)√2]²=[1...

Given the preceding term and Sn = ((an + 1) & # 178;) / 4 of positive number sequence {an}, find the general term formula of positive number sequence {an}

A1 = S1 = (a1 + 1) ^ 2 / 4, so A1 = 1
an=sn-s(n-1)
(an-1)^2=(a(n-1)+1)^2
Because it is a positive number, so an-1 = a (n-1) + 1
an=a(n-1)+2
Arithmetic sequence
an=1+2(n-1)=2n-1

The general term formula of sequence {an} is an = n & # 178; - 10N + 1 (1) find the first three terms (2) judge whether 25 is one of the terms (3) find the minimum value of the sequence

（1）a1=1²-10×1+1=-8
a2=2²-10×2+1=-15
a3=3²-10×3+1=-20
(2) Let an = n & # 178; - 10N + 1 = 25
Then n & # 178; - 10n-24 = 0
(n-12)(n+2)=0
The solution is n = 12 or n = - 2 (rounding off)
So 25 is one of them
（3）an=n²-10n+1=(n-5)²+1-25=(n-5)²-24≥-24
So item 5 is the minimum, and the minimum is - 24