Prove the following by mathematical induction:
$4^n > n+1$, for all integers $n ≥ 1$
Step 1:
$n=1$:
LHS $= 4^{(1)} = 4 $
RHS $= (1) + 1 = 2$
LHS > RHS. ∴ $P(1)$ is true.
Step 2:
Assume $P(k)$ is true for some $k ≥ 1$
$P(k)$: $4^k > k+1$
Step 3: We must show $P(k+1)$ is true.
$n = k+1$: $4^{k+1} > (k+1)+1$
RHS = $(k+1)+1 < 4^k +1 < 4^k + 4^k = 2*4^k < 4*4^k = 4^{k+1}$ = LHS
Hence, $P(k+1)$ is true. Therefore, By Math. Induction $P(n)$ is true for all $n ≥ 1$
Can anyone check if my method is correct or there is a better way to do it. Thank you.
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