Prove using the principle of mathematical induction that
(1+x)^n> (or equal to) 1+nx, n=1,2,3...
thank you !
This is very straightforward.
If $\displaystyle (1+x)^n\ \ge\ 1+nx$ ......(1)
then $\displaystyle (1+x)^{n+1}\ must\ be\ \ge\ 1+(n+1)x$ ....... (2)
Try to prove this using (1)
as this means (1) being true for some k causes (1) to be true for all n >k.
$\displaystyle (1+x)^k(1+x)\ \ge\ (1+kx)(1+x)$ ?
$\displaystyle (1+x)^{k+1}\ \ge\ 1+kx+x+kx^2$ ?
$\displaystyle (1+x)^{k+1}\ \ge\ 1+(k+1)x+kx^2$ which is > $\displaystyle 1+(k+1)x$