Question

Mathematical induction of 3k-1 ≥ 4k ( 3k = k power of 3 )

Answer 1

For making mathematical induction, we need:

• a base case

An $n_0$ for which the relation holds true

• the induction step

if its true for $n_i$, then, is true for $n_{i+1}$

base case

the relationship is not true for 1 or 2

$1^3-1 = 0 < 4*1$

$2^3-1 = 8 -1 = 7 < 4*2 = 8$

but, is true for 3

$3^3-1 = 27 -1 = 26 > 4*3 = 12$

induction step

lets say that the relationship is true for n, this is

$n^3 -1 \ge 4 n$

lets add 4 on each side, this is

$n^3 -1 + 4 \ge 4 n + 4$

$n^3 + 3 \ge 4 (n + 1)$

now

$(n+1)^3 = n^3 +3 n^2 + 3 n + 1$

$(n+1)^3 \ge n^3 + 3 n$

if $n \ge 1$ then $3 n \ge 3$ , so

$(n+1)^3 \ge n^3 + 3 n \ge n^3 + 3$

$(n+1)^3 \ge n^3 + 3 \ge 4 (n + 1)$

$(n+1)^3 \ge 4 (n + 1)$

and this is what we were looking for!

So, for any natural equal or greater than 3, the relationship is true.