Finite sequence 1,3,5 The number of terms of 2n-3 is_____

Finite sequence 1,3,5 The number of terms of 2n-3 is_____

1=2*1-1
3=2*2-1
5=2*3-1
2n-3=2*(n-1)-1
So the number of terms is n-1

What is the number of terms of finite sequence 1,3,5,..., 2n-3

1 = 2n-3 (n = 2) the number of items is 1
3 = 2n-3 (n = 3) the number of items is 2
5 = 2n-3 (n = 4) the number of items is 3
……
2n-3 is item n-1
So it's n-1

8 out of 21 divided by 16 out of 7 divided by 26 left 13 out of 9 answers

8/21÷16/7÷26×13/9
=(8/21×7/16)×(1/26×13/9)
=1/6×1/18
=1/108

Teacher, I've come to trouble you! It's a question of fraction
It's just one (x-4 / 1 + X + 4 / 1) / / x square - 16 / 2. I don't know how to find the most common denominator. How to find the most common denominator for this kind of problem?

Solution
simple form
=[(x+4)/(x-4)(x+4)+(x-4)/(x-4)(x+4)]÷2/(x-4)(x+4)
=[(x+4+x-4)/(x+4)(x-4)]÷2/(x-4)(x+4)
=(2x)/(x-4)(x+4)×(x-4)(x+4)/2
=x
I don't know how to ask

How do you find the least common denominator of a fraction?

Look for his quality factor

Try to explain that the square difference of two continuous positive even numbers must be divisible by 4, but not by 6

Two consecutive positive even numbers can be expressed as X and X + 2
The square difference of two numbers can be expressed as: (x + 2) ^ 2-x ^ 2 = x ^ 2 + 4x + 4-x ^ 2 = 4x + 4 = 4 (x + 1)
So it must be divisible by 4
In addition, there are errors in the last half of the title, and the square difference may be divided by 6
For example, when x = 2, 2 ^ 2 = 4, 4 ^ 2 = 16, 16-4 = 12, 12 can be divisible by 6

By means of factorization, it is proved that the square difference of two continuous even numbers must be divisible by 4

It is proved that if any even number is 2n and N is an integer, then another even number is 2n + 2 and N is an integer
The difference between the squares of two consecutive even numbers is
(2n) ^ 2 - (2n + 2) ^ 2 factorization
The original formula = (2n + 2n + 2) (2n - (2n + 2))
=-2*(4n+2)
=-4(2n+1)
Because the original formula can be decomposed into the product of 4 and another factor, it must be divisible by 4

Can the square difference of two continuous even numbers be divisible by 4? Why?

Let two continuous even numbers be 2n, 2n + 2, then (2n + 2) 2 - (2n) 2, = (2n + 2 + 2n) (2n + 2-2n), = (4N + 2) × 2, = 4 (2n + 1). Since n is an integer, 2n + 1 in 4 (2n + 1) is also a positive integer, so 4 (2n + 1) is a multiple of 4

After experiments, Xiao Ming found a rule that the square difference of two consecutive even numbers must be divisible by 4. Please prove it

Let the smaller even number be a, then the larger even number be (a + 2)
（a+2）²-a²
=a²+4a+4-a²
=4a+4
=4（a+1）
The square difference of a continuous even number must be divisible by 4

The sum of the squares of three consecutive positive integers is 77. Find the three numbers

These three numbers are 4, 5 and 6