The sum of two consecutive numbers is S. The square of the larger number minus the square of the smaller number is?
let the two numbers be x and x+1
x + x+1 = S
2x = S - 1
x = (S-1)/2
The square of the larger number minus the square of the smaller number
= (x+1)^2 - x^2
= x^2 + 2x + 1 - x^2
= 2x + 1
= (S-1) + 1
= S
check:
let's take 37, and 38
sum = 75
square of larger =1444
square of smaller = 1369
1444 - 1369 = 75 , which is the sum of the 2 numbers.
This can also be seen by noting that
a^2-b^2 = (a+b)(a-b)
In this case, since the numbers are consecutive, that is just
S*1 = S
To solve this problem, we need to understand the relationship between consecutive numbers and squares.
Let's assume the smaller of the two consecutive numbers is x. This means that the larger consecutive number would be x+1, as it comes after the smaller number.
The sum of these two consecutive numbers is given as S. So, we can write the equation as:
x + (x+1) = S
Simplifying this equation, we get:
2x + 1 = S
Now, let's calculate the squares of the larger and smaller numbers:
Square of the larger number = (x+1)^2 = x^2 + 2x + 1
Square of the smaller number = x^2
To find the difference between the squares, we subtract the smaller square from the larger square:
(x^2 + 2x + 1) - x^2 = 2x + 1
So, the difference between the squares of the consecutive numbers is 2x + 1.