Find Patterns 7th grade, 7th grade math find patterns test questions

n*(n+2)*(n+4)*(n+6)+16n*(n+6)*(n+2)*(n+4)+16(n^2+6n)(n^2+6n+8)+16(n^2+6n)[(n^2+6n)+8]+16(n^2+6n)^2+8(n^2+6n)+16(n^2+6n+4)^2

The number is a perfect square.

n=2k n^2+6n+4)^2

4k^2+12k+4)^2

16(k^2+3k+1)^2

The result is a perfect multiple of 16.

This is fill in the blanks: 2 4 6 8 + 16 = 400

Regularity: n*(n+2)*(n+4)*(n+6)+16n*(n+6)*(n+2)*(n+4)+16(n 2+6n)(n 2+6n+8)+16(n 2+6n)[(n 2+6n)+8]+16(n 2+6n) 2+8(n 2+6n)+16(n 2+6n+4) 2

The number is a perfect square.

n=2k n^2+6n+4)^2

4k^2+12k+4)^2

16(k^2+3k+1)^2

The result is a perfect multiple of 16.

1，4，9，16，25...are all squared numbers, i.e. n, and the 100th is 100 = 10000. The 20th is 20 = 400.

2，-8，-18，-32，-50……is -2n and the 20th digit is -2*20 = -800.

1，4，9，16，25……How is the law of this set of numbers expressed, and what are its 100th and 20th numbers?

The n2 20th digit is 20 2=400

The 100th number is 100 2 = 10000

2，-8，-18，-32，-50……How to express the law of this group of numbers, and what is its 20th number?

The 20th digit of 2*n 2 is -2*20 2=-800

The 100th number is -2*100 2=-20000

1 squared, 2 0 squared, 3 squared. The 100th number is 10000 squared by 100, and the 20th number is 20 squared.

The next number is smaller than the previous number, 2+(n-1)*4, and the general formula for the nth term = -2+(-2+(-4))+2+2*(-4))+

There are a total of n-2 in it; The latter can be seen as a series of equal differences, and the formula for summing the first term is 0, the tolerance is -4, and the result is -2*n*(n-1).

Substituting the above equation, the nth term = -2*n + (-2) * n * (n-1) = -2n squared.

1. If all digits are squared, 100 numbers are 10000, and 20 numbers are 400

2. The opposite number of 20 is -400*2=-800

Hope that helps you :)

1 2 2 2 3 2 4 2 5 2 is squared The 100th number 100 2 = 10000 The 20th number 20 2 = 400

Each is n squared, the hundredth is 10,000, and the 20th is 400,

n+1)^2

When n is 0, the value is 1, and n is a natural number, which satisfies the condition. Thank you.

6*6=—36—；13*13=169——；15*15=—225—；

1) Write down the patterns you discovered. The square of a number is equal to the sum of the product of its two adjacent numbers and 1.

2) According to the above rules, try to calculate the root number under 2011*2013+1=2012——; 2011*2013+1=(2012-1)(2012+1)+1=2012 squared, so 2011*2013+1=2012 under the root number

The last null should not be this n(n+1)=——。 under the root number.

Rule: x*x=(x-1)(x+1)+(1 squared), fill in the blanks, the first 2012, the second is, the last question is not set correctly,

The square of n (n+1)(n-1)-1

No, I misspelled the square of n (n+1)(n-1)+1

So the result is 2012 and don't know the latter one.

The last of the nth line is n(n+1) 2, so the last of the 10th line is 10 11 2 55

The first of line nth is n(n+1) 2-n+1, so the first of line 20 is: 20 21 2 20 1 191

Its rule is very simple, the first of each horizontal line is the first number of the previous row plus the ordinal number of the row, for example, 7=4+3, 3 is the ordinal number of the line starting with 4. In the line, it is a continuous positive integer. The number of numbers in each row is equal to the ordinal number of rows.

The last digit in each row is equal to the total number of digits in that row.

So that the last digit in line 10 is equal to (1+10)10 2=55

A number in line 20 is 19 (1+19) 2+1 = 191

The last number in the tenth row is 55, and the first number in the 20th row is 191

Observe a set of monomials: x, 3x x x, 9x x...What is the 2013th formula?

The coefficient is odd, x is the number is 1, 2, 3,。。

So the 2013th equation is: (2 2013-1)x to the power of 2013 = 4025x to the power of 2013;

Observe the following monomials: a, -2a, 4a, -8a to the fourth power. What is the pattern? What is the eighth formula?

The coefficient is a proportional series, and the common ratio is -2, i.e., 1, -2, 4, -8, 16, -32, 64, -128 ,..

The times were: 1, 2, 3 ,..

So the eighth equation is: -128a to the 8th power;

The square of a, thirds (the fourth power of a), the fifth (the sixth power of a), and the seventh (the eighth power of a), then the nth equation is?

The coefficient is the reciprocal of the odd number: i.e. the coefficient of the nth equation is 2n-1/1

The number of times is: 2, 4, 6 ,..Even column: 2n

So the nth equation is: (2n-1) to the 2n power of a;

3 matches in one triangle, 5 matches in two triangles, 7 matches in three triangles, how many matches in n triangles?

are odd numbers: 3, 5, 7, 9 ,..arrangement, so n triangles to: 2n + 1 match;

8 matches for one goldfish, 14 matches for two goldfish, 20 matches for three goldfish, how many matches for n goldfish? Expressed in integers.

The first term is 8, and the tolerance is 6 for a series of numbers, an=8+(n-1) 6=6n+2

So the nth article requires: 6n + 2 matches.

My bounty points are gone, please.

x^3；（"^"denotes the superscript, i.e., "power") (2), -128a 8;The rule of this problem is: the first term: (-2) 0*a 1;The second item:

2)^1*a^2；Each item is: (-2) (n-1)*a n;

3), (a(2n)) (2n-1), i.e., (2n-1) ((2n) power of a);

n+1；(Regularity).

n+2；(Regularity).

If you need to know more details, please ask!

（1+n）*n/2

The nth digit is 1+2+......The sum of n.

You can do the math with a series of equal differences.

If you haven't learned, you can use the following methods to calculate!

1+ 2 +…n-1)+n

n+(n-1)+…2 +1

Just add the above two formulas and divide them by two.

It is understood as: the sum corresponding to the addition of upper and lower numbers is (n+1), there are n such sums, and then divided by 2 is the result of the formula.

Find the law:

The nth number is 1+2+3+4+·· n = n(n+1) 2Note: In the last step, the formula for summing the difference series is used.

Solution: Observations:

The first number is equal to: 1=1

The second digit is equal to: 1+2=3

The third digit is equal to: 1+2+3=6

The fourth digit is equal to: 1 + 2 + 3 + 4 = 10

The nth number is equal to: 1+2+3+4+5... n is calculated to be equal to: [n*(n+1)] 2

(1) 12 20 The first one is 1, the second one is 2, and the third one is the sum of the first two plus 1 (2) -48 96 to the nth power.

3)125 216 n³

4) 31 63 2 n times minus 1