Law exploration: 5,12,21,32 The nth

Law exploration: 5,12,21,32 The nth

(n+2)*(n+2)/(n+2)*(n+2)-4

1,3,11,43,171…… N observe and find the law, summarize the formula

Observe and find the law
A (n + 1) - an contains 4 ^ n
Let an = a + B4 ^ n
a1=a+4b=1
a2=a+16b=3
---> a=1/3,b=1/6
an=(1/3)+(1/6)4^n

A question about equal difference and equal ratio sequence. Urgent!
It is known that an is an arithmetic sequence, and A2 is equal to 1, A5 is equal to - 5

a5=a2+3d
So d = - 2 A1 = a2-d = 3
Sn=na1+[n(n-1)d]/2=-n^2+4n
Let's look at the function f (x) = - x ^ 2 + 4x
When x = 2, take the maximum,
When Sn = 2, the maximum value is 4

The product of four consecutive odd numbers is 19305. What are the four odd numbers

By decomposing 19305 into the mass factor, we get the following results:
19305=3*3*3*5*11*13
So 19305 = 9 * 11 * 13 * 15

Let p be the product of any three adjacent positive odd numbers not exceeding 1987, then what is the largest integer that can divide all such p

There are six ways to satisfy such P
1.1*3*5
2.3*5*7
3.5*7*9
4.7*9*11
5.9*11*13
6.11*13*15
（13*15*17 >1987）
From the above six numbers, the largest integer divisible is 3

The product of three adjacent odd numbers is 1 □ □ 7. The three odd numbers are______ ．

Since the end of the product is 7, there must be 1 or main, 3 or 9 at the end of the three odd numbers. These three odd numbers are adjacent: because 1 and 7 are not adjacent odd numbers, 3 and 9 can be adjacent, that is, 9, 11, 13, 19, 21, 23, 29, 31, 33 It is verified that: 9 × 11 × 13 = 1287. Meet the requirements. That is, the three numbers are: 9, 11, 13. So the answer is: 9, 11, 13

Try to explain: the product of three consecutive integers can be divided by 6

Three consecutive integers
At least one of them must be even, so the product can be divisible by 2
And at least one of them is a multiple of 3, so it can also be divisible by 3

The sum of two different positive integers which are not 3 is a multiple of 3. The sum of the product of the two integers plus 1 can be divided by 3,
For example: 15 = 7 + 8, 7 * 8 + 1 = 57, 57 is a multiple of 3
6 = 2 + 4 2 * 4 + 1 = 9 9 is a multiple of 3
27 = 7 + 20 7 * 20 + 1 = 141 141 is a multiple of 3
If you can prove it, if you can't, please give a counter example!

Let these two numbers be x, y have: x + y = 3k, K is a positive integer,
xy+1
=x(3k-x)+1
=3kx-x^2+1
=3kx-(x+1)(x-1)
Because: 3kx can be divisible by 3, and (x + 1) (x-1) can also be divisible by 3,
So the sum of the product of two numbers plus one can be divided by three

The sum of three consecutive positive integers must be divisible by 3. The sum of three consecutive even integers must be divisible by 6
These are two questions

Let these three positive integers be n, N + 1 and N + 2
Then n + N + 1 + N + 2 = 3N + 3 = 3 (n + 1)
Because 3 (n + 1) is divisible by 3, the sum of three consecutive positive integers must be divisible by 3
Let three consecutive even numbers be 2x, 2x + 2, 2x + 4
Then 2x + 2x + 2 + 2x + 4 = 6x + 6 = 6 (x + 1)
Because 6 (x + 1) is divisible by 6, the sum of three consecutive even numbers must be divisible by 6

If the product of the first 2011 positive integers is 1 × 2 × X 2011 can be divided by 2010k, then the maximum value of positive integer k is______ ．

∵ 2010 = 2 × 3 × 5 × 67, ∵ the largest number after decomposition is 67, ∵ starting from 67, then 67 × 1 Up to 67 × 30, a total of 30, the maximum can only be 30. The maximum value of positive integer k is 30