How do I prove that (n+5)^100 = big-theta(n^100)?

I know that the highest degree is 100 and it equals to the degree of n on the right, which proves that (n+5)^100 = big-theta(n^100).

I was just wondering if there is another way to prove this with work instead of just this theorem.

  • 2
    $\begingroup$ One can use the fact that for $n>5$ we have $(n+5)^{100}<(2n)^{100}=2^{100}n^{100}$. $\endgroup$ – Wojowu Feb 8 '15 at 21:40
  • $\begingroup$ See math notation guide. $\endgroup$ – user147263 Feb 9 '15 at 0:11

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