Prove $5^n + 3^n - 2^{2n+1} > 0$ by induction

Question

I am not sure how to deal with the $-2^{2n+1}$ term.

I did the basis proof for n=1

I am stuck at this step: $$ 5^{k+1}+3^{k+1}-2^{2(k+1)+1} = 5\cdot 5^k + 3 \cdot 3^k -2^3 \cdot 2^{2k} $$

Any advice guys?

Answer 1

HINT

Note that $2^{2n+1} = 2^{2n} \cdot 2 = 2 \cdot 4^n$. So you have to prove that $$5^n + 3^n > 2 \cdot 4^n.$$

You claim you did the base case. Now assume this is true for all integers upto $n$ and prove that $$5^{n+1} + 3^{n+1} > 2 \cdot 4^{n+1}.$$

Answer 2

Hint: $5^n > 2^{2n+1}$ for $n \ge \ldots$.

Answer 3

$5(5^k)+3(3^k)-2^3(2^{2k})=3(5^k+3^k-2^{2k+1})+2(5^k)-2(2^{2k})>3(0)+2(5^5-4^k)>0$

Answer 4

The base case is simple. We now assume $5^k+3^k-2^{2k+1}>0$. Then multiplying by 4 yields $4\cdot 5^k + 4 \cdot 3^k - 4\cdot 2^{2k+1}>0$, which implies $5\cdot 5^k + 3\cdot 3^k -2^{2k+3} > 5^k - 3^k >0$ as required.

Answer 5

Let $a_n:=5^n+3^n-2\times 4^n$ so $a_0=0,\,a_{n+1}-5a_n=2(4^n-3^n)\ge 0$. Induction gives $a_n\ge0$ viz. $a_{n+1}\ge 5a_n\ge 0$. The inequality becomes strict at $n=2$, viz. $a_2=2$.

Answer 6

Note that $$ \begin{align} 5^{k+1}+3^{k+1}-2^{2(k+1)+1} &=5\cdot5^k+3\cdot3^k-4\cdot2^{2k+1}\\ &=\left(5^k-3^k\right)+4\left(5^k+3^k-2^{2k+1}\right) \end{align} $$