How much is the price of a 1980 yuan TV that is sold by a shopping mall at a 20% discount and still makes a profit of 10%? Lie equation~

How much is the price of a 1980 yuan TV that is sold by a shopping mall at a 20% discount and still makes a profit of 10%? Lie equation~

Set the price as X, 0.8 * x = 1980 * (1 + 10%)

In order to enliven the economy, a shopping mall will sell a commodity a at a 10% discount on the price, and still make a profit of 10%. If the price of commodity a is 33 yuan, then the purchase price is ()
A. 31 yuan B. 30.2 yuan C. 29.7 yuan D. 27 yuan

Let the price of the goods be x yuan. Then according to the meaning of the question, we can get: (1 + 10%) x = 33 × 90%, the solution is: x = 27, so we choose D

Who can teach me: cumulative method, cumulative addition, column elimination method Wait. (there can be a better way, better!

There are many methods in sequence,
For example, a (n + 1) / an = 1 / n,
A (n + 1) - an = n,
Split term elimination method, 1 / [n (n + 1)] = 1 / N - 1 / (n + 1)
When subtracting by dislocation and summating by equal ratio sequence,
The undetermined coefficient method is usually used when the general term formula is derived from the recursive formula

It is known that the sum of the first n terms of an is Sn, and Sn is equal to 1 / 3 (an-1)
Find A1, A2

A1=S1=1/3(A1-1)
3A1=A1-1
A1=-1/2
S2=A1+A2=1/3(A2-1)
-3/2+3A2=A2-1
A2=-1/4

TN-1 * TN + 1 = TN * TN + 5 is converted into a recurrence formula, and T1 = 1, T2 = 2 are known
Note: where n-1, N + 1 and N are subscripts

t(n-1)*t(n+1)=tn*tn+5
When n = 2, T1 * T3 = (T2) ^ 2 + 5, T3 = 9
When n = 3, T2 * T4 = (T3) ^ 2 + 5, T4 = 43
(tn)^2-t(n-1)t(n+1)+5=0
[t(n-1)]^2-t(n-2)tn+5=0
Subtraction of two formulas:
(tn)^2-[t(n-1)]^2=t(n-1)t(n+1)-t(n-2)tn
(tn)^2+t(n-2)tn=[t(n-1)]^2+t(n-1)t(n+1)
[tn+(1/2)t(n-2)]^2=[t(n-1)+(1/2)t(n+1)]^2
tn+(1/2)t(n-2)=±[t(n-1)+(1/2)t(n+1)]
t(n+2)+(1/2)tn=±[t(n+1)+(1/2)t(n+3)]
one
When + is taken: t (n + 2) + (1 / 2) TN = t (n + 1) + (1 / 2) t (n + 3)
t(n+2)-t(n+1)=(1/2)t(n+3)-(1/2)tn
=(1/2)[t(n+3)-tn]
=(1/2)[t(n+3)-t(n+2)+t(n+2)-t(n+1)+t(n+1)-tn]
=(1/2)[t(n+3)-t(n+2)]+(1/2)[t(n+2)-t(n+1)]+(1/2)[t(n+1)-tn]
Let an = t (n + 1) - TN, A1 = t2-t1 = 1, A2 = t3-t2 = 7
a(n+1)=(1/2)a(n+2)+(1/2)a(n+1)+(1/2)an
a(n+1)=a(n+2)+an
a(n+2)-a(n+1)+an=0
Let a (n + 2) - Xa (n + 1) = y [a (n + 1) - xan]
x+y=1,xy=1
x1=(1+√5)/2,y1=(1-√5)/2
x2=(1-√5)/2,y2=(1+√5)/2
a(n+2)-xa(n+1)=[a2-xa1]y^n=(7-x)y^n
Let [a (n + 2) + Z] - x [a (n + 1) + Z] = 0
a(n+2)-xa(n+1)=xz-z
z=[(7-x)y^n]/(x-1)
[a(n+2)+z]=x[a(n+1)+z]
a(n+2)+z=(a1+z)x^(n-1)
a(n+2)=-z+(a1+z)x^(n-1)
an=-z+(1+z)x^(n-3)=t(n+1)-tn
t(n+1)-tn=-z+(1+z)x^(n-3)
=-[(7-x)y^n]/(x-1)+{1+[(7-x)y^n]/(x-1)}x^(n-3)
tn-t(n-1)=
t(n-1)-t(n-2)=
……
t2-t1=
The results are as follows
two
Take - time: t (n + 2) + (1 / 2) TN = - t (n + 1) - (1 / 2) t (n + 3)

The master of series
In the sequence an. A1 = 1, an + an + 1 = 3 of the nth power. Find the sum of the first n terms. I got this sequence by myself. It's 1,2,7,20,61. The rule is to multiply 3 and then subtract 1 or add 1. Who can help me do it? It needs a certain process

If it's not easy to make subscripts, you might as well remember that an is the first n items of a (n) and Sn
2A (n) = 3 ^ n - 1 (1)
2A(n-1)=3^(n-1)-1 （2）
.
.
.
2A(2)=3*3-1
2(A1)=3-1
Add all the above equations to get
2Sn=3+3*3+.+3^(n)-n
That is 2S (n) = (3 ^ (n + 1) - 3) / 2 - n
Sn = (3^(n+1)-3-2n)/4
Top: S1 = 1
S2=5
S3=18
Experience is right!

Let tn be the product of the first n terms of the sequence {an}, satisfying TN = 1-an (n is a positive integer)
【1】 Let BN = 1 / an, prove that the sequence {BN} is an arithmetic sequence, and find an and BN

Sorry, we begin to regard TN as the sum of the first n terms of the sequence {an}. Now correct, TN = 1-an, t (n + 1) = 1-A (n + 1) a (n + 1) = t (n + 1) / TN = [1-an] / [1-A (n + 1)] and get: 1 / [1-A (n + 1)] - 1 / [1-an] = 1, let BN = 1 / [1-an], then: B (n + 1) - BN = 1, so BN is the arithmetic sequence A1 = T1 = 1-a1 a

Application of mathematical sequence
One asks for a free fall from a height of 100 meters. After landing each time, the ball bounces back to half of the original height and then falls. How many meters has the ball passed when it lands for the 10th time? (reserve to one rank)
Merry Christmas

First, consider the path of falling, A1 = 100a2 = 50. Q = 1 / 2n = 10, sum formula of equal ratio sequence, Sn = A1 (1-Q ^ n) / (1-Q) S10 = 100 * (1-1 / 2 ^ 10) / (1-1 / 2) ≈ 199.8m, then consider the rising, A1 = 50a2 = 25... Q = 1 / 2n = 9, the 10th time, S9 = 50 * (1-1 / 2 ^ 9) / (1-1 / 2) ≈ 99.8m

There is a column of numbers: A1, A2, A3 If A1 = 2, then a2007 is ()
A. 2007B. 2C. 12D. -1

According to the meaning of the question: A1 = 2, A2 = 1-12 = 12, A3 = 1-2 = - 1, A4 = 1 + 1 = 2; period is 3; 2007 △ 3 = 669; so a2007 = A3 = - 1

Two questions
1： The sum of the first n terms of the sequence {an} is Sn = 2An + 3n-12, and BN = an * n. find the sum of the first n terms of {BN} and TN
2： A (n + 2) = 5 / 3A (n + 1) + 2 / 3an,
Find the first n terms and TN of {n * an}
Thank you very much

1,Sn-S(n-1)=an=2an-2a(n-1)+3
∴an-3=2[a(n-1)-3]
The common ratio is equal to 2, A1 = S1 = 2A1 + 3-12, A1 = 9
∴an=3+6*2^(n-1)
∴bn=3n+6n*2^(n-1)
BN is divided into two parts 3N and 6N * 2 ^ (n-1) to sum respectively
I can't write it down. I'll leave you a message offline