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Question:

Prove $ \ \forall n \ge 2, \ 2^{n} + 3^{n} < 4^{n}$

My attempt:

Base case is trivial.

Suppose $ \ n \ge 2$ and $ \ 2^{n} + 3^{n} < 4^{n}$

Then,

$2^{n+1} + 3^{n+1} = 2.2^{n} + 3.3^{n} = 2.2^{n} + 2.3^{n} + 3^{n} = 2(2^{n} + 3^{n}) + 3^{n} <2(4^{n}) + 3^{n} $, by I.H.

I am stuck here. how do I show that this expression is $ < 4^{n+1}$?

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  • $\begingroup$ Hints: Is $3^n<4^n$? Is $3(4^n)<4(4^n)$? $\endgroup$ – Michael Burr Sep 16 '17 at 20:25
  • $\begingroup$ Another proof could use that the equivalent inequality $(2/4)^n+(3/4)^n<1$ after dividing both sides by $4^n$, where $4^n>0$, holds because it holds for $n=2$ and $(2/4)^n$ and $(3/4)^n$ are both strictly decreasing. $\endgroup$ – user236182 Sep 16 '17 at 20:32
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You can say $$2\cdot 2^n + 3\cdot 3^n < 4\cdot 2^n + 4\cdot 3^n = 4(2^n+3^n) < 4\cdot 4^n = 4^{n+1}.$$

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multiplying $$2^n+3^n<4^n$$ by $$4^n$$ we get $$2^n\cdot 2^{2n}+3^n\cdot 2^{2n}<4^{n+1}$$ and we have $$2^n\cdot 2<2^{3n}$$ since $$1<2^{2n-1}$$ for $n\geq 2$ and $$3^n\cdot 3<3^n\cdot 2^{2n}$$ since $$3<2^{2n}$$ therefore we have $$2^{n+1}+3^{n+1}<2^n\cdot 2^{2n}+3^n\cdot 2^{2n}<4^{n+1}$$

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Hint:   for $\,0 \lt a \lt 1\,$ the sequence $\,a^n\,$ is decreasing with $\,n\,$ since $\,a^{n+1} = a \cdot a^n \lt 1 \cdot a^n\,$, so:

$$\left(\frac{2}{4}\right)^n+\left(\frac{3}{4}\right)^n \le \left(\frac{2}{4}\right)^2+\left(\frac{3}{4}\right)^2 = \frac{13}{16} \lt 1 \quad\quad\text{for}\;\; \forall n \ge 2$$

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