Question #202811

3n>2n2




1
Expert's answer
2021-07-28T17:48:08-0400

We want to prove by induction that


3n>2n23^n > 2n^2

Base case: n=1;


31>2(1)2    3>23^1 > 2(1)^2 \implies 3>2

Thus S1S_1 is true.


Inductive Step:

We assume it is true for n=k, so that


3k>2k23^k > 2k^2

Inductive Step:

We will show that since it is true for n= k, then it is true for n=k+1.

3k+1=3k3>2k23>6k2>2k2+4k2>2[(k+1)22k1]+4k2>2(k+1)24k2+4k23k+1>2(k+1)2+4k24k23^{k+1} = 3^k \cdot 3\\ \qquad > 2k^2 \cdot 3\\ \qquad > 6k^2\\ \qquad > 2k^2 + 4k^2\\ \qquad > 2[(k+1)^2-2k-1] + 4k^2\\ \qquad > 2(k+1)^2 -4k -2 + 4k^2\\ 3^{k+1} > 2(k+1)^2 +4k^2-4k-2\\


Since


3k+1>2(k+1)2+4k24k23^{k+1} > 2(k+1)^2 +4k^2-4k-2\\

then


3k+1>2(k+1)23^{k+1} > 2(k+1)^2

which concludes the proof.


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