Question

If P (n) is the statement "2ⁿ ≥ 3n" and if P (r) is true, prove that P (r + 1) is true.

Answer 1

We have:
$$\begin{gathered} P\left(n\right):2^n\ge 3n \\ \mathrm{Also}, \\ P\left(r\right)\mathrm{is}\mathrm{true}. \\ \therefore 2^r\ge 3r \\ \mathrm{To}\mathrm{Prove}:P(r+1)\mathrm{is}\mathrm{true}. \\ Wehave: \\ {2}^r\ge 3r \\ \Rightarrow 2^r\times 2\ge 3r\times 2\left[\mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}2\right] \\ \Rightarrow 2^{r+1}\ge 6r \\ \therefore 2^{r+1}\ge 3r+3\left[6r\ge 3r+3\mathrm{for}\mathrm{every}r\in N.\right] \\ \mathrm{Hence},P(r+1)\mathrm{is}\mathrm{true}. \end{gathered}$$