Prove that $2^n\geq2n-1$ Let $n\geq2$ with $n\in\Bbb N$. Prove that
$$2^n\geq2n-1$$
I need to prove this using mathematical induction. This is what I've tried:
$P(2): 2^2\geq2n-1 \\
P(k)\Rightarrow P(k+1) \\
P(k+1): 2^{k+1}\geq2k+1 \\
\begin{align}
2\cdot2^k & \geq2k -1+2 \\
2^k & \geq2k-1
\end{align}
$
I am not sure if what I've done will finally lead to something or if it's already enough, but please, tell me how can I handle this type of exercises in general! Thank you!
 A: It is much better writing first your assumption and then write your deductions. That is: first write $P(k)$:
$$2^k\ge2k-1$$
Now try to do something to get $P(k+1)$. In this case, add $2^k$:
$$2^k+2^k\ge2k-1+2^k$$
which is$$2^{k+1}\ge 2k-1+2^k$$
Now use the fact $2^k>2$:
$$2^{k+1}\ge2k-1+2^k>2k-1+2=2(k+1)-1$$
and you are done.
Alternatively, you can try to write $2^{k+1}$ and make a chain of inequalities that ends in $2k+1$, using in this process that $P(k)$ which you have assumed:
$$2^{k+1}=2\cdot2^k\stackrel{P(k)\text{ used here}}\ge2(2k-1)=(2k-1)+(2k-1)\ge(2k-1)+2$$
A: $\displaylines{
  k = 1 \Rightarrow 2 \ge 1\quad true \cr 
  if\quad 1 < k:2^k  \ge 2k - 1 \cr 
   \Rightarrow 2 \times 2^k  \ge 2 \times \left( {2k - 1} \right) \cr 
   \Rightarrow 2^{k + 1}  \ge 4k - 2 \ge 2k + 1 \cr 
   \cr 
  if:4k - 2 < 2k + 1 \Leftrightarrow 2k - 3 < 0 \cr 
   \Leftrightarrow k < \frac{3}{2} \Rightarrow k = 1\quad or\;k = 0 \cr}$
A: we have to prove that $2^{n+1}\geq 2(n+1)-1$ if $2^n\geq 2n-1$ is hold, multiplying the last inequality by $2$ we get
$2^{n+1}\geq 2(2n-1)=4n-2\geq 2n+1$
A: What you did is almost correct, the one nitpick is that you want to prove that $P(k)\implies P(k+1)$ and then you start with the fact that $P(k+1)\implies2^{k+1}\ge 2k+1$ and arrive at $P(k+1)\implies 2^k\ge 2k-1$ which is the same as $P(k+1)\implies P(k)$. However the thing you're trying to prove is $P(k)\implies P(k+1)$ so you need to start with $2^k\ge 2k-1$.
