The difference of their squares is:

(n+k)^2 - n^2

Expanding (n+k)^2, we have:

(n+k)^2 = n^2 + 2nk + k^2

Substituting this back into the expression for the difference of squares, we get:

(n+k)^2 - n^2 = (n^2 + 2nk + k^2) - n^2

Simplifying, the n^2 terms cancel out:

(n+k)^2 - n^2 = 2nk + k^2

Next, we need to show that this expression is equal in magnitude to k times the sum of n and n+k.

The sum of n and n+k is:

n + (n+k) = 2n + k

Now, k times their sum is:

k(2n + k)

Expanding this, we have:

k(2n + k) = 2nk + k^2

We observe that:

(n+k)^2 - n^2 = 2nk + k^2 = k(2n + k)

Thus, the difference of the squares of n and n+k is indeed equal to k times their sum. This completes the proof.

Problem Bank - Proof and Reasoning