Urgent master solution on a series of general formula problem! The sequence {an} satisfies a1 + 2A2 + 3a3 + Nan = n (n + 1) (n + 2), find the general term formula of sequence {an} Please help write a detailed problem-solving process, thank you very much! If you can, please help to write the solution to similar problems in the future I'm very grateful (0.0)!

Let BN = n * an
The sum of BN is TN
Then a1 + 2A2 + 3a3 + nan=Tn=n(n+1)(n+2)
Then t (n-1) = (n-1) n (n + 1)
bn=Tn-T(n-1)=n(n+1)(n+2)-(n-1)n(n+1)
=n(n+1)(n+2-n+1)
=3n(n+1)
So an = BN / N = 3N + 3
Let n = 1, A1 = 1 * 2 * 3 = 6;
According to the meaning of the title:
a1+2a2+3a3+....+nan=n(n+1)(n+2)，
a1+2a2+3a3+....+nan+(n+1)a(n+1)=(n+1)(n+2)(n+3)
By subtracting the two formulas, we get (n + 1) a (n + 1) = (n + 1) (n + 2) (n + 3) - n (n + 1) (n + 2), so a (n + 1) = (n + 2) (n + 3) - n (n + 2) = 3... Expansion
Let n = 1, A1 = 1 * 2 * 3 = 6;
According to the meaning of the title:
a1+2a2+3a3+....+nan=n(n+1)(n+2)，
a1+2a2+3a3+....+nan+(n+1)a(n+1)=(n+1)(n+2)(n+3)
By subtracting the two formulas, we get (n + 1) a (n + 1) = (n + 1) (n + 2) (n + 3) - n (n + 1) (n + 2), so a (n + 1) = (n + 2) (n + 3) - n (n + 2) = 3 (n + 2),
Thus: an = 3 (n + 1) (n ≥ 2)
An (a ≥ 2) is also suitable for A1
So the general term an = 3 (n + 1)
If the line passing through point (1,0) is tangent to the curve y = x & # 179; and y = ax & # 178; + 15 / 4x-9, the value range of real number a is obtained
The derivative of y = x ^ 3 is y = 3x ^ 2. If the tangent point of the line is (m, m ^ 3), then the line passes through (m, m ^ 3), (1,0), then the line is y = 0 or y = 27 / 4 * (x-1) if y = 0. Then the vertex of y = ax ^ 2 + 15 / 4x-9 is a = - 25 / 64. If y = 27 / 4 * (x-1), the slope is 27 / 4Y = ax ^ 2 + 15 / 4x-9, and the derivative of y = 2aX + 15 / 4, then the line and
Series 2,1,1 / 2 general term formula series 3 / 2,9 / 4,65 / 16 general term formula how to solve this kind of problem in detail
In general, through observation, subtraction, quotient, decomposition, expansion and other methods, 1) subtraction: A1 = 1A2 = 2A3 = 4a4 = 7... Subtraction found a2-a1 = 1a3-a2 = 2a4-a3 = 3 natural conjecture: an-an-1 = n-1 and then add up to an-a1 = (1 + 2 + 3 +... + n-1) (as long as the sum can be solved) 2) quotient (with the title example): A2 / A1 = 1 /
2,1,() ,1/2
Written as: 2 / 1, 2 / 2, (2 / 3), 2 / 4
It can be written in the form of fractions at a glance
an=2/n
It should be: 3 / 2, 9 / 4, 25 / 8, 65 / 16
3/2=1+1/2
9/4=2+1/4
25/8=3+1/8
65/16=4+1/16
.....
Therefore:
An =... Expand
2,1,() ,1/2
Written as: 2 / 1, 2 / 2, (2 / 3), 2 / 4
It can be written in the form of fractions at a glance
an=2/n
It should be: 3 / 2, 9 / 4, 25 / 8, 65 / 16
3/2=1+1/2
9/4=2+1/4
25/8=3+1/8
65/16=4+1/16
.....
Therefore:
an=n+(1/2)^(n)
This kind of problem is based on observation, memorizing arithmetic sequence, proportional sequence, harmonic sequence and other basic sequence, in addition to more questions
1. The general formula is 2 ^ (2-N)
2. The general formula is n + (1 / 2) ^ n
Mainly observation.
In the first group of sequence, the latter is 1 / 2 of the former, so the general term formula is obtained
In the second group of sequences, it is observed that the denominator is a multiple of 2, so we try to change it into the form with fraction
They are: 1 + 1 / 2, 2 + 1 / 4, 4 + 1 / 16
So we get the general formula.
If there is a straight line passing through the point (1,0) and the curves y = x ^ 3 and y = ax ^ 2 + 15x / 4-9 are tangent, find the value of the real number a
Firstly, the slope k of the straight line L passing through point (1,0) is obtained by the condition that the line L is tangent to the curve C1: y = x ^ 3. Let the equation of the straight line l be y = K (x-1), where k is the slope, and let the tangent point of L and C1 be a (x0, Y0). Since point a is not only on L but also on C1, two numerical relations between x0 and Y0 can be obtained: Y0 = K (x0 - 1) ① Y0 = x0 ^ 3
A very difficult sequence problem! Find a recursive sequence to find the general term formula!
A（n+1）= m*An+Bn
Where BN general term formula is known, BN = n * q ^ n-1, q is a constant
Find the general term formula of an!
BN = n * q ^ (n-1) q is a constant
[1] In this paper, we obtain the following result: if Q (n) is not equal to anq (1 + n) / N, then q (n) = anq (1 + n) / N ^ (a + 1) / N}
The best way to solve this problem is not to use mathematical induction, because there are too many unknowns to induce.
(undetermined coefficient method)
You don't give A1, so let A1 = a
A（n+1）= m*An+n*q^(n-1)
When m ≠ 1,
Let a (n + 1) + xnq ^ (n-1) = m (an + xnq ^ (n-1))
The solution is x = 1 / (m-1)
So let CN = an + NQ ^ (n-1) / (m-1),
Where C1 = a + 1 / (M -... Expansion
The best way to solve this problem is not to use mathematical induction, because there are too many unknowns to induce.
(undetermined coefficient method)
You don't give A1, so let A1 = a
A（n+1）= m*An+n*q^(n-1)
When m ≠ 1,
Let a (n + 1) + xnq ^ (n-1) = m (an + xnq ^ (n-1))
The solution is x = 1 / (m-1)
So let CN = an + NQ ^ (n-1) / (m-1),
Where C1 = a + 1 / (m-1),
It can be obtained that {CN} is C1 = a + 1 / (m-1), and the common ratio is m
So CN = (a + 1 / (m-1)) * m ^ (n-1)
So an + NQ ^ (n-1) / (m-1) = (a + 1 / (m-1)) * m ^ (n-1)
Simplify
An=(a+1/(m-1))*m^(n-1)-nq^(n-1)/(m-1)
When m = 1, we can add up
You have to tell A1
Using the method of undetermined coefficient, we can get the
（An+？） Is equal North sequence
Let a (n + 1) + D = t * (an + D)
Then we can find the general term of the new sequence, and then we can find the original general term
It's difficult to calculate because of the complexity of calculation. Besides, it's hard to type
The second floor is also a good way
The most widely used sequence is construction and mathematical induction
A(n+1)=mAn+Bn
A(n+1)+cq=m(An+c)
We obtain MC CQ = BN = n * q ^ (n-1)
c=Bn/(m-q)
Dn=An+Bn/(m-q)
Then d (n + 1) = m * DN
Dn=D1*m^(n-1)=(A1+1)*m^(n-1)
An=Dn-Bn/(m-q)=(A1+1)*m^(n-1)-n*q^(n-1)/(m-q)
The title does not give A1. If there is A1, we can find the general term
Guess before prove, mathematical induction
If there is a line passing through the point (1,0) and the curves y = x ^ 3 and y = ax ^ 2 + 3x-9 are tangent, find the value of the real number a
Let the line passing through point (1,0) be tangent to the curve y = x ^ 3
The slope is K and the tangent point is m (x0, Y0)
For y = x & # 179; derivation, y '= 3x & # 178;,
∴k=3x²0; ①
y0=x³0 ②
k=y0/(x0-1) ③
Eliminate K, Y0: 3x & # 178; 0 = x & # 179; 0 / (x0-1)
The solution is x0 = 3 / 2, k = 27 / 4
The tangent equation is as follows:
y=27/4(x-1)
Derivation of y = ax ^ 2 + 3x-9
y'=2ax+3
Let tangent point n (x1, Y1)
Then 2ax1 + 3 = k = 27 / 4 = > ax1 = 15 / 8
y1=27/4(x1-1)
y1=ax²1+3x1-9
Eliminate Y1, K
∴27/4(x1-1)=15/8*x1+3x1-9
∴54x1-54=15x+24x1-72
∴x1=-6/5
∴a(-6/5)=15/8
∴ a=-25/16
How to find the first five terms of the general term formula with a sequence problem 3 / 2 2 / 3 5 / 12 3 / 10 7 / 30
4/21
Change 2 / 3 to 4 / 6 and 3 / 10 to 6 / 20
The numerator is 3 4 5 6 7 8 and the denominator is 1 * 2,2 * 3,3 * 4,4 * 5 * 6 6 * 7
So 8 / 42 = 4 / 21
How to solve the equation of 3 * (2.4-0.75x) + 4x = 9.3?
3*(2.4-0.75x)+4x=9.3
3*(2.4-0.75x)=9.3-4x
(2.4-0.75x)=3.1-4x/3
7x/12=0.7
x=1.2
Hope to adopt
The general term formula of sequence
1.0.7、0.77、0.777、0.7777-----
2.√3、3√15、√21、3√3----
3.-1＼5 3＼10 －5＼17 7＼26
Find a n respectively=
The first: (10 ^ n-1) * 7 / 90, the second you play right, the third (- 1) ^ n * [2N-1 / (n + 1) ^ 2 + 1]
1、a1=0.7
a2=0.7+0.7*0.1
……
an=0.7[1+0.1+0.1^2+…… +0.1^(n-1)]=0.7*[(1-0.1^n)/0.9]
=7/9*[1-(1/10）^n]
4X + 9 * (2x / 3) = 24? How to solve this equation
4x+18x÷3=24
10x=24
x=2.4
4x+9*(2x/3)=24
4x + 3*2x = 24
4x + 6x = 24
10x =24
x = 2.4

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