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Question

For every positive integer $$n, \dfrac {n^7}{7}+\dfrac {n^5}{5}+\dfrac {2n^3}{3}-\dfrac {n}{105}$$ is


A
an integer
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B
a rational number
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C
a negative real number
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D
an odd integer
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Solution

The correct option is A an integer
$$\dfrac { { n }^{ 7 } }{ 7 } +\dfrac { { n }^{ 5 } }{ 5 } +\dfrac { 2{ n }^{ 3 } }{ 3 } -\dfrac { n }{ 105 } $$ 
where $$n$$ is a positive integer.
When $$n=1$$, we get,
$$\dfrac { { n }^{ 7 } }{ 7 } +\dfrac { { n }^{ 5 } }{ 5 } +\dfrac { 2{ n }^{ 3 } }{ 3 } -\dfrac { n }{ 105 } $$ 
$$=\dfrac { 15{ n }^{ 7 }+21{ n }^{ 5 }+70{ n }^{ 3 }-n }{ 105 } $$
$$=\dfrac { 15+21+70-1 }{ 105 } =\dfrac { 105 }{ 105 } =1$$ which is an integer.
When $$n=2$$,
we get;
$$\dfrac { { n }^{ 7 } }{ 7 } +\dfrac { { n }^{ 5 } }{ 5 } +\dfrac { 2{ n }^{ 3 } }{ 3 } -\dfrac { n }{ 105 } $$ 
$$=\dfrac { 15{ n }^{ 7 }+21{ n }^{ 5 }+70{ n }^{ 3 }-n }{ 105 } $$
$$=\dfrac { \left( 15\times 128 \right) +\left( 21\times 32 \right) +\left( 70\times 28 \right) -2 }{ 105 } $$
$$=\dfrac { 1920+672+560-2 }{ 105 } $$
$$=\dfrac { 3105 }{ 105 } =30$$ which is an even positive integer.


Mathematics

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