In the above table you can see that each set of consecutive numbers created shows the “Middle number squared” is +1 more than the “First number multiplied by the last”.
From this pattern I will solve it by using algebra. The equation for the “Chosen consecutive numbers” is: n, n + 1, n + 2.
Formula
n, n + 1, n + 2
= (n + 1)² - n(n + 2)
= (n + 1)(n + 1) - (n² + 2n)
= n² + n + n + 1 - (n² + 2n)
= n² + 2n + 1 - (n² + 2n)
= 1
Test the formula
I will test my formula to prove it is correct, by replacing “n” with one of the first consecutive numbers:
When “n” = 9
= (9 + 1)² - 9(9 + 2)
= (9 + 1)(9 + 1) - (9² + 2 x 9)
= 9² + 9 + 9 + 1 - (9² + 2 x 9)
= 9² + 2 x 9 + 1 - (9² + 2 x 9)
= 1
The equation above shows that my formula for this “Problem” is correct.
I will now do “Problem 2” and see if I can find a rule in algebra for consecutive numbers.
Problem 2
From this table you can see that “Differences between squared numbers” column is the same number I get when I add the two numbers in the “Chosen numbers” column.
From this pattern I will try to find a rule in algebra.
Formula
n, n + 1
= n² - (n + 1)²
= n² - (n + 1)( n + 1)
= n² - n² + n + n +1
= n² - n² + 2n + 1
= 2n + 1
Test the formula
To test the above algebra and to prove I am correct I must now test the formula. My formula for this problem is “2n + 1”, I must replace “n” with one of the consecutive numbers I have chosen. For example: if n = 5, the sum will look like this: 2 x 5 + 1 = 11.
I will test my formula in this table:
As the above table shows my formula is correct. The answers I get from this table match my earlier table correctly.
I have now done “Problem 2” but there is no clear pattern I can see, so I will find more rules for different consecutive patterns.
Problem 3
In this “Problem” I have chosen to use 5 consecutive numbers. I have squared the middle number and multiplied the first number by the fifth to see if there is a clear pattern, and as you can see the difference between the “Middle number squared” and the “First number multiplied by the fifth” columns is 4.
Formula
n, n +1, n + 2, n + 3, n+ 4
= (n + 2)² - n² + 4n + 4
= n(n + 4) - n² + 4n
= n² + 4n + 4 - (n² + 4n) = 4
Testing the formula
I will now test my formula to see if it’s correct. I will replace “n” with “5” and if it is correct the answer will be 4.
When n = 5
= (5 + 2)² - 5² + 4 x 5 + 4
= 5(5 + 4) - 5² + 4 x 5
= 5² + 4 x 5 + 4 - (5² + 4 x 5) = 4
As you can see my formula is correct and I am starting to see a pattern in the consecutive number patterns, and I predict that the pattern is: (Gap between consecutive numbers)².
I will continue with the “Problems” to prove my Prediction is correct.
Problem 4
The above table shows a Problem with numbers in consecutive order but with a gap between them, e.g. 1, 3, 5.
Formula
n, n + 2, n + 4
= n(n + 4) - (n + 2)²
= n² + 4n - (n + 2)(n + 2)
= n² + 2n + 2n + 4 - n² + 4n
= n² + 4n + 4 - n² + 4n
= 4
Testing the formula
When n = 8
= 8(8 + 4) - (8 + 2)²
= 8² + 4 x 8 - (8 + 2)(8 +2)
= 8² + 2 x 8 + 2 x 8 + 4 - 8² + 4 x 8
= 8² + 4 x 8 + 4 - 8² + 4 x 8
= 4
I can now see it is highly likely my prediction was correct and the pattern being: (Gap between consecutive numbers)².
__________________________________________________________________________
Working out the formula
The final formula for working out the pattern in consecutive numbers is shown below:
n, n + g, n + 2g
= n(n +2g) + (n + g)²
= n(n +2g) + (n + g) (n + g)
= (n² + 2gn) + (n² + gn + gn +g²)
= (n² + 2gn + g²) – (n² + 2gn)
= g²
With the above formula I can workout any consecutive number sum, I replace “n” with the number(s) used and change “g” with the amount of the gap between each number.
E.g. if the consecutive numbers chose where: 2, 4, 6 the “g” in the formula would be replaced with 2, because that’s the size of the gap between numbers in this case.
