Asked by Janine
The sum of two consecutive numbers is S. The square of the larger number minus the square of the smaller number is?
Answers
Answered by
Reiny
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.
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.
Answered by
Steve
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
a^2-b^2 = (a+b)(a-b)
In this case, since the numbers are consecutive, that is just
S*1 = S
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