How to solve the equation (a + 3 + a) x2 = 52?

How to solve the equation (a + 3 + a) x2 = 52?

(2a+3)x2=52
2a+3=26
2a=23
a=23/2

(3 + x) x2 = 9 solution equation

（3+X)x2=9
（3+X)=9÷2
3+X=4.5
X=4.5-3
X=1.5

Solving equation x2-1 = x

x2-1=x
x²-x=1
x²-x+1/4=1+1/4
(x-1/2)²=5/4
x-1/2=±√5 /2
x=1/2±√5 /2
Click comment in the upper right corner, and then you can select satisfied, the problem has been solved perfectly

It is known that the first n terms of sequence {an} and Sn = - n ^ 2 + 9N + 2, n belongs to n*
(1) Judge whether {an} is arithmetic sequence
(2) Let RN = | A1 | + | A2 | + +|An |, find RN
(3) Let BN = 1 / [n (12-an)], n belong to n *, TN = B1 + B2 + +BN, whether there is a minimum natural number N0, such that the inequality TN

1、a(n)=Sn-S(n-1) = -n^2+9n+2-(-(n-1)^2+9(n-1)+2) n>=2
=10-2n
a1 = S1 = 10
An is not an arithmetic sequence
2. When n > = 5 | an | = 2n-10
When n = 5, RN = - Sn + 2 * S4 = - n ^ 2 + 9N + 46
3、bn = 1/[n*(2n+2)] = (1/2)*[1/n - 1/(n+1)]
Tn = (1/2)*[1/1 - 1/2 + 1/2 -1/3 +…… +1/n - 1/(n+1)]
= (1/2)*[1-1/(n+1)]
Tn (1/2)*[1-1/(n+1)] < (n0/32)
1-1/(n+1) < n0/16
In order to make n always true for all non-zero natural numbers, we need only:
1-1/2 < n0/16
n0 > 8
There is N0, and the minimum is 9

The first n terms and Sn = n ^ 2-9n-1 of known sequence an
1. Find the general term of an
2, find the sum of the first n terms of {an}

For reference only! (1) an = Sn - (sn-1) = n & # 178; - 9n-1 - [(n-1) &# 178; - 9 (n-1) - 1] = 2n-10  an general term is: an = 2n-10 (2) let an ≥ 0 get: n ≥ 5

Given the first n terms of sequence an and Sn = - N & # 178; + 9N + 8, (1) find the general term formula an of sequence an
(2) If A1 and A3 are the third term and the third term of the proportional sequence BN respectively, find the first n term and TN of the sequence BN

1. We can know that the sum of the first n-1 terms is sn-1 = - (n-1) ^ 2 + 9 (n-1) + 8
an=sn-sn-1=-n²+9n+8-(-(n-1)^2+9(n-1)+8)=-2n+10
2. Your topic is incomplete and can't be worked out

The first N-term sum formula of known sequence {an} is Sn = n & # 178; - 9N + C (C belongs to R)
(1) If the sequence {an} is an arithmetic sequence, find the value of C
(2) Find the value of serial number n when Sn gets the minimum value

(1)
Sn=n^2-9n+c
n=1
a1= -8+c
for n>=2
an = Sn-S(n-1)
= 2n-1-9
= 2n-10
a1= -8+c
2-10=-8+c
c=0
(2)
2n-10 >0
n> 5
min Sn = S4 or S5

The first n terms and Sn = nxn-9n of sequence {an} are known, and the k-th term satisfies 5

Sn=nxn-9n (n=1 2 3 4 .)
S(n+1)=(n+1)x(n+1)-9x(n+1)
=nxn-7n+10
a(n+1)=S(n+1)-Sn
=2n-8
an=2n-10
It is concluded that an = 2 (N-5)
So think 5

Given Sn = 1 / 8 (an + 2) ^ 2, prove: {an} is arithmetic sequence

a1=S1=(1/8)(a1+2)^2 (a1-2)^2=0 a1=2
Sn=(1/8)(an+2)^2
S(n-1)=(1/8)(a(n-1)+2)^2
Subtraction of two formulas
8[S(n)-S(n-1)]=(an+2)^2-(a(n-1)+2)^2
(a(n-1)+2)^2=(an+2)^2-8an
(a(n-1)+2)^2=(an-2)^2
When a (n-1) + 2 = - (An-2)
a(n-1)+a(n)=0
Rounding off the contradiction with positive integer column {an}
When a (n-1) + 2 = An-2
an-a(n-1)=4
Then {an} is an arithmetic sequence an = a1 + (n-1) d = 2 + 4 (n-1) = 4n-2

1 / (n-1) n (n + 1) how to split the term? It is mainly used to find the number sequence

an = 1/[(n-1)n(n+1)] ; n ≥2
=(1/2){ 1/[(n-1)n] -1/[n(n+1)] }
a2+a3+...+an = (1/2) { 1/(1.2) - 1/[n(n+1)] }