Question

Prove that (n+5)^2-(n+3)^2 is a multiple of 4 for all positive integer values of n

Answer 1

Answer:

Proved

Step-by-step explanation:

Given

$(n+5)^2-(n+3)^2$

Required

Show that $(n+5)^2-(n+3)^2$ is a multiple of 4

Expand each bracket

$(n+5)(n+5)-(n+3)(n+3)$

Open brackets

$n^2 + 5x + 5x + 25 - (n^2 + 3x + 3x + 9)$

$n^2 + 10x + 25 - (n^2 + 6x + 9)$

Open bracket

$n^2 + 10x + 25 - n^2 - 6x - 9$

Collect Like Terms

$- n^2 + n^2 - 6n + 10n + 25 - 9$

$- 6n + 10n + 25 - 9$

$4n + 25 - 9$

$4n + 16$

Factorize

$4(n + 4)$

Hence, the multiples of $(n+5)^2-(n+3)^2$ are $4$ and $(n + 4)$