Proof by induction when numbers are to powers

Question

Prove by mathematical induction: $$ 2^n+3^n < 5^n$$

Answer 1

The given inequality is valid for $n=2$. Assume that it is valid for some $n$, $n\ge 2$. Then for $n+1$ we have

$$5^{n+1}=5\times 5^n>5(2^n+3^n)>2\times 2^n+3\times 3^n=2^{n+1}+3^{n+1}$$ and we get the result.

Answer 2

since we have $$\left(\frac{2}{5}\right)^n+\left(\frac{3}{5}\right)^n<1$$ it is clear for all $n>1$.

Answer 3

Show $2^2+3^2<5^2$

Assume $2^n+3^n<5^n$

Prove $2^{n+1}+3^{n+1}<5^{n+1}$:

• $2^{n+1}+3^{n+1}=2\cdot2^n+3\cdot3^n$
• $2\cdot2^n+3\cdot3^n<3\cdot2^n+3\cdot3^n$
• $3\cdot2^n+3\cdot3^n=3\cdot(2^n+3^n)$
• $3\cdot(2^n+3^n)<3\cdot5^n$ assumption used here
• $3\cdot5^n<5\cdot5^n$
• $5\cdot5^n=5^{n+1}$

Answer 4

This is a proof without induction:
Lets define two continuous functions $f,g:\mathbb{R}^+_0\to\mathbb{R}$ such that $$f(x)=\Big(\dfrac{2}{5}\Big)^x$$ and $$g(x)=1-\Big(\dfrac{3}{5}\Big)^x$$ Note that $f(0)=1$ and $g(0)=0.$
Also $f$ is strictly decreasing and $g$ is strictly increasing.
Therefore they can have at most one intersection point. They intersect for $x=1$: $f(1)=\dfrac 2 5 = g(1)$. So, in particular, $g(n) > f(n)$ for all $n \geq 2$. That is to say $1-3^n/5^n > 2^n/5^n$ or $5^n > 2^n + 3^n$ for all $n \geq 2$.

Answer 5

By the binomial theorem, $5^n=(2+3)^n=2^n+\cdots+3^n > 2^n+3^n$, because all missing terms are positive. Note that you need $n\ge2$ to have any missing terms.