Solve an equal difference ratio problem. How to solve this analogy problem?

Let an=a1+(n-1)d, a1 be the first term, and d be the tolerance, so that sn=n*a1+(n*(n-1) 2)*d, then sn n=a1+((n-1) 2)*d is substituted into n:

Equispecific properties: (a1+d)*(a1+, introduce a1=(-5 3)d or d=0 equal properties: (a1+d)+(a1+

From the above two equations, we can deduce a1=4, d=-12 5=d=0, and a1=1

sn is the sum of the first n terms of the difference series an, then.

s3/3=a2

s4/4=( a2+a3 )/2

s5/5=a3

Now, s3 3*s4 4=s5 5*s5 5 i.e. a2*( a2+a3 ) 2=a3*a3 a2*a2+a2*a3=2a3*a3 1

s3 3+s4 4=1*2=2 i.e. a2+( a2+a3 ) 2=2 3a2+a3=4 2

Combining the above two equations, we can calculate a2 and a3

Calculate two answers, 1, a2=, a3= an=

2、a2=a3=1 an=1

First of all, neither a1 nor 1-q is 0. Divide both sides of the equation by a1 (1-q) and simplify:

1 - q³ +3(1 - q²) 0

1-q)(1+q+q²) 3(1-q)(1+q) =01-q)(1+q+q² -3 - 3q) =01-q)(q²-2q-2) =0

So, q -2q - 2 = 0

q²-2q+1 = 3

q-1)² 3

So, q - 1 = 3

Then the wide and bright in the fierce:

q = 3 + 1, or q = 1 - 3

Featuring a brick.

The first time: with a 2+1 to leave a 2-1

The second time: use (a 2-1) 2+1=a 4+1 and leave a 4-3 2

The third time: use (a 4-2 3) 2+1=a 8+1 and leave a 8-7 4

A 2 9+1 was used for the ninth time

So sn=a 2+a 4+a 8+......a/2^9+(9)=a

The 9th floor used up more than half of the remaining piece of the 8th floor, which happened to be used up, so the remaining pieces of the 8th floor were two;

The 8th layer uses up more than half of the remaining pieces of the 7th layer, so the remaining half of the 7th layer is less than 1, that is, two pieces, and the 7th layer is pushed out and there are 6 pieces left;

From this reasoning, there are a total of 1022 bricks.

Of course, it can also be considered in the method of equal difference series.

Let the number remaining for the nth time be an, then there is the recursive formula an=(an-1) 2-1

Make sure you don't make a mistake in the question.

Equal Difference Series Proportional Series Sum of the first k terms.

a(1) = a1 b(1) = b1

a(2) = a1 + d b(2) = b1*d

a(3) = a1 + 2d b(3) = b1*d²

a(n) = a1 + n-1)d b(n) = b1*d^(n-1)

According to the title: a1=b1

a(4) = b(4) ===> a1 + 3d = b1*d³ ②

a(10)=b(10) ===> a1 + 9d = b1*d^9 ③

Simultaneous solution: a1 = 2 (1 3) d = - 2 (1 3).

Or a1 = 0 d = 0 -- I wonder if the current textbook should be discarded? It's up to you.

b(16) = a1*d^15 = -2^(16/3)

Suppose there is an item a(k+1) = b(16).

i.e. 2 (1 3) (1 + k) = - 2 (16 3).

then k +1 = - 2 5

k = - 33 is not a natural number, so b16 is not an term in an.

There are many similarities between these two series, in fact, the proportional series is a difference series after taking the logarithm. It is helpful to keep the following points in mind:

1.Both are two unknowns, the first term A1 and the tolerance (ratio) q, need two conditional columns and two equations to solve.

is the difference (quotient) of two adjacent terms

3.Either term is the arithmetic (geometric) average of the two terms:

That is, the equal difference series: 2an=(an-1)+a(n+1), and the proportional series an 2=a(n-1)a(n+1).

4.Sum by the first term and the tolerance (ratio): na1+n(n-1)q 2; a1[1-q^(n-1)]/(1-q)

5.Sum by the first and last terms: (a1+an)n 2, a1[1-q (n-1)] (1-q), q=(an a1) [1 (n-1)].

6.The odd term is summed by n times (to the power) of the intermediate term am: nam, (am) n

7.Even terms are summed by n 2 times (power) of the middle terms am, am+1 and (product): (am+am+1)n 2, (amam+1) (n 2).

Because a, b, c are proportional sequences.

So b= (ac).

So b 2 = ac

The two sides are logarithmic.

lgb^2=lg(ac)

It can be obtained according to the nature of logarithms.

2lgb=lga+lgc

So LGA, LGB, LGC are equal difference series.

If b>a>1, then the difference column is 1,a,b, and the ratio column is 1,a,b, then 1+b=2a,a2=b

1+a 2=2a, a=1, b=1, not eligible, rounded;

If b>1>a, then the difference is listed as a,1,b, and the proportional number is listed as a,1,b or 1,a,b

At this time, a+b=2, ab=1, the solution gives a=1 - root number 2, b = 1 + root number 2 or a + b = 2, b = a 2, solution gives a = -2, b = 4 If 1 > b > a, then the difference is listed as a, b, 1, and the ratio is listed as a, 1, b or 1, b, a

a+1=2b,ab=1

At this time, a, b have no real solutions;

or a+1=2b, b 2=a, in this case a=b=1, rounded.

So there are only two sets of solutions: a=a=1-root number 2, b=1 + root number 2 or a=-2, b=4