Question 801968
Let {{{ 2n }}} = the 1st even integer
{{{ 2n + 2 }}} = the 2nd even integer
{{{ 2n + 4 }}} = the 3rd even integer
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given:
{{{ 2n + 2n + 4 = 2*( 2n + 2 ) }}}
{{{ 4n + 4 = 4n + 4 }}}
{{{ n }}} can be any integer, so {{{ 2n }}}
can be any even integer
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No matter which 3 consecutive even integers
you choose, like 12, 14, 16
the 1st plus the 3rd = twice the 2nd
{{{ 12 + 16 = 2*14 }}}
{{{ 28 = 28 }}}
OK