Show that $(3n + 1)^2 + 2$ is always divisible by 3 for all integer values of $n$ . Answer

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Worked solution

1. Expand and simplify the given expression.

$(3n+1)^2+2=9n^2+6n+1+2$
$=9n^2+6n+3$

2. Factorise the expression to show it is a multiple of 3.

$9n^2+6n+3=3(3n^2+2n+1)$
Since $n$ is an integer, $3n^2+2n+1$ is also an integer. Hence the expression is divisible by 3.

Answer

$(3n+1)^2+2=3(3n^2+2n+1)$
Since $3n^2+2n+1$ is an integer for all integer values of $n$ , $(3n+1)^2+2$ is always divisible by 3.

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