Solve a mathematical problem with a sequence of numbers, a problem with a series of numbers

You're talking about a new sequence of numbers that are multiplied by the equivalence of the difference series and the corresponding of the proportional series, (the reciprocal sequence of the proportional series is also the proportional series, so you said that.)"than", which can be seen as multiplication) is commonly known"Difference. than"Series, with"Misplaced subtraction"Summation, that is, the method of deriving the first n terms and formulas of the proportional number series in the textbook. Page 61 in the textbook (Renjiao Society A), exercise 4

3) Questions are also of this kind. --i.e. sn= first"Difference. than"The first n terms of the series are written as sums, placed in the upper row, and then multiplied by the common ratio of 1 bn on both sides of the equation, that is, 1 qsn=...

Put down the row, wrong one place on the right, so that the number of times the common ratio is aligned with the upper row, and then subtract the two formulas ,..Generally, the result of subtraction can be summed with a proportional series. Study and understand the two questions in the above textbook first, and you will be done.

Why is your question so vague ... What are you looking for? Ask for an sn? And what is the general formula for an and bn?

In fact, it is also a proportional difference series, and you will know it after you have scored, and the sum of the proportional difference series should be found, right?

First, find the nth term of an than bn and calculate it directly with their general term.

The 10th place will receive a prize of A1 and the 9th place will receive a prize of A2,..The prize money for the first place is A10, and when there is no score, it is A1 = 2, A2 = 2 (A1 + 1), A3 = 2 (A2 + 1), .,a(n+1)=2(an +1)

So a(n+1) +2=2(an +2) is a series of proportional 2, so that an +2=a1 ·2 (n-1)=2 n

an =2^n -2

Thus a11 = 2 11 -2 = 20.46 million yuan.

Reverse thrust 10th place to take 2w left

9th place is left before (2+1) 2=6w

8th place takes the remaining (6+1) 2=14w, and so on.

A total of 2046w was taken out

What kind of unit is this, I really want to go to o( o

To add: The money left for the nth time is all 2 to the n+1 power minus 2

Let's look at the pattern first, the numbers are all arranged in order, and each row is four digits except for the first row.

So the number of rows where 79 is located should be counted like this:

79-5=74 (minus the five numbers in the first row.)

74 4 = 18 and 2 (divide the remaining number by the number of numbers in each row to get the number of rows, and the remaining 2 means that there are two more numbers, i.e. 73 and 74 in the next row).

So the number of rows where 74 is located = 1 + 18 + 1 = 20

In the same way, the number of rows where 68 is located:

Hope it helps, thank you.

There are 5 in the first row and 4 in the back.

74 4 = 18 remainder 2

then not counting the first row, 79 is in the 18th row.

Plus the first row.

So 79 in the 19th row.

Again, 68-5=63

63 4 = 15 and 3

So 68 in the 16th row.

The relative number difference between each other line is 8, and you will know it anyway.

Finding the maximum value of the number of quadratic letters, sn=3n -16n=3(n -16 3+64 9)-64 3=3(n-8 3) -64 3 When n=8 3, sn is the smallest, because n is the number of the whole book, so the closest is n=3, and when this posture is potato, sn=-21

a1=1=3^0=3^(1-1)

a2=3=3^1=3^(2-1)

a3=9=3^2=3^(3-1)

a4=27=3^3=3^(4-1)

an=3^(n-1)

The general formula for the series is an=3 (n-1).

An is a proportional series, 8*A2+A5=0, A5=A2*Q (5-2)=A2*Q 3, 8*A2+A2*Q 3=0, Q=-2 ,A1=-(A2) 2

a5 a3= a2*q 3 a2*q=-8 -2=4s5 s3= a1*=[q (n+1)-1] [q (n)-1] cannot find the value.

The value that cannot be found is d s(n+1) sn

Solution: an=4 n-2 n, and the two sides are summed by the sigma symbol of successive positive integers from 1 to n (the left is sn, and the right is the difference between the two proportional sequences).

sn=4 3(4 n -1)-2(2 n-1)bn=2 n sn=2 n than 4 3(4 n -1)-2(2 n-1) The right side can be arranged into 3 2(1 (2 n-1) -1 (2 (n+1)-1)).

i.e. bn 3 2(1 (2 n-1) -1 (2 (n+1)-1)) then tn=3 2(1-1 3+1 3-..1/(2^(n+1)-1））=3/2(1-1/(2^(n+1)-1) )

a(n)=4^n-2^n

s(n)-s(n-1)=4 n-2 n(s(n),s(n-1) represent the sum of the first n terms and the first n-1 terms of the series a, respectively

s(n-1)-s(n-2)=4^(n-1)-2^(n-1)

s(2)-s(1)=4 2-2 2 The left and right sides are added to give :

s(n)-s(1)=(4^2+··4^(n-1)+4^n)-(2^2+··2^(n-1)+2^n)

s(1)=a(1)=2

s(n)=s(1)+(4^2+··4^(n-1)+4^n)-(2^2+··2^(n-1)+2^n)=2+4^2(4^(n-1)-1)/(4-1)-2^2(2^(n-1)-1)

S(n) is found by the proportional sequence formula, b(n) is obtained by b(n)=2 n s(n), and t(n) is found by a similar method as s(n).

This method seems to be more cumbersome, and the lack of feasibility when finding t(n) later, if you want to find the first n terms of 1 b(n) and can also (although it is very troublesome), ah I can't do it, but I wrote a lot, or submitted it, I hope there is a master to make it