Show that $(2n-1)^2 + 3$ is a multiple of 4 for all integer values of $n$ . Answer ... ...

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Worked solution

1. Expand and simplify the expression.

$(2n-1)^2+3=(2n-1)(2n-1)+3$
$=4n^2-4n+1+3$
$=4n^2-4n+4$

2. Factorise out 4 and justify that the remaining factor is an integer.

$4n^2-4n+4=4(n^2-n+1)$
Since $n$ is an integer, $n^2-n+1$ is also an integer.

Therefore $(2n-1)^2+3$ is a multiple of $4$ for all integer values of $n$ .

Answer

$(2n-1)^2+3=4(n^2-n+1)$
Since $n$ is an integer, $n^2-n+1$ is an integer. Hence $(2n-1)^2+3$ is a multiple of $4$ .

AI analysis — may contain mistakes.

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