Visitor

2015-12-09 12:47:22

#1

Find out the naturals *n* for which we have

2^{3n+1} < (n+4)^{2}

2

Member

2015-12-09 13:04:24

#2

This could be an answer.

For n=0

2^{1} < 4^{2} .... and this is true because 2 < 16

For n=1

2^{4} < 5^{2} ..... and this is true, as well, because 16 < 25

For n=2

2^{7} < 6^{2} .... but here it is not true anymore because 256 > 36

For n =3

210 < 72 ... neither here is not true

....

We can see that the first member increase faster then the second one so we won't have another solution.

so though: S={0, 1}

For n=0

2

For n=1

2

For n=2

2

For n =3

210 < 72 ... neither here is not true

....

We can see that the first member increase faster then the second one so we won't have another solution.

so though: S={0, 1}

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