1) Start with the left-hand side: $\frac{1}{4}(2n+1)(n+4) - \frac{1}{4}n(2n+1)$.
Factor out the common term $\frac{1}{4}(2n+1)$:
$\frac{1}{4}(2n+1)[(n+4) - n]$.
Simplify inside the brackets: $(n+4) - n = 4$.
So, the expression becomes $\frac{1}{4}(2n+1)(4) = (2n+1)$.
This equals the right-hand side, so the proof is complete.
2) Expand both squares:
$(4n + 2)^2 = 16n^2 + 16n + 4$.
$(2n + 2)^2 = 4n^2 + 8n + 4$.
Subtract them: $(16n^2 + 16n + 4) - (4n^2 + 8n + 4) = 12n^2 + 8n$.
Factor out 4: $4(3n^2 + 2n)$.
Since $3n^2 + 2n$ is an integer for any positive integer $n$, the expression is a multiple of 4.
3) Expand both squares:
$(n + 1)^2 = n^2 + 2n + 1$.
$(n - 1)^2 = n^2 - 2n + 1$.
Substitute into the expression: $(n^2 + 2n + 1) - (n^2 - 2n + 1) + 1$.
Simplify: $n^2 + 2n + 1 - n^2 + 2n - 1 + 1 = 4n + 1$.
For any positive integer $n$, $4n$ is even, so $4n + 1$ is odd.
4) Let the two consecutive even numbers be $2n$ and $2n+2$, where $n$ is a positive integer.
Their squares are $(2n)^2 = 4n^2$ and $(2n+2)^2 = 4n^2 + 8n + 4$.
The difference is $(4n^2 + 8n + 4) - 4n^2 = 8n + 4$.
Factor out 4: $4(2n + 1)$.
Since $2n + 1$ is an integer, the difference is a multiple of 4.
5) Let the two numbers be $a$ and $b$.
The product of their difference and sum is $(a - b)(a + b)$.
By the difference of squares formula, this equals $a^2 - b^2$.
This is exactly the difference of their squares, so the statement is proven.
6) Apply the Pythagorean theorem to the right triangle: $(6x)^2 + (x + y)^2 = (8x - y)^2$.
Expand each term: $36x^2 + (x^2 + 2xy + y^2) = 64x^2 - 16xy + y^2$.
Combine like terms on the left: $37x^2 + 2xy + y^2 = 64x^2 - 16xy + y^2$.
Subtract $y^2$ from both sides: $37x^2 + 2xy = 64x^2 - 16xy$.
Bring all terms to one side: $0 = 64x^2 - 16xy - 37x^2 - 2xy$.
Simplify: $0 = 27x^2 - 18xy$.
Assuming $x \neq 0$, divide by $9x$: $0 = 3x - 2y$.
Rearrange: $3x = 2y$, or $\frac{x}{y} = \frac{2}{3}$.
Thus, $x:y = 2:3$.
Parent Tip: Review the logic above to help your child master the concept of algebraic proofs worksheet with answers.
