Question 1022193
Assume induction hypothesis true for n = k, where *[tex \large k \ge 1]:

*[tex \large 3^k \ge 1 + 2k]

You want to prove that:

*[tex \large 3^{k+1} \ge 1 + 2(k+1) = 3 + 2k]


First, you can note that *[tex \large 3^{k+1} = 3 \cdot 3^k], and this is greater than or equal to *[tex \large 3(1+2k)] by induction hypothesis. So

*[tex \large 3^{k+1} \ge 3+6k]


We also know that *[tex \large 3+6k \ge 3+2k] since *[tex \large k \ge 1].


Therefore


*[tex \large 3^{k+1} \ge 3+6k \ge 3+2k]


So the statement is true for n = k+1.