Question

If $a={99}^{100}$, $b={100}^{99}$, then

• A. $a>b$
• B. $a=b$
• C. $a<b$
• D. $a+2b={101}^{100}$

Answer 1

The correct option is A $a>b$

$\frac{{100}^{99}}{{99}^{99}}$
$=\frac{(1+99{)}^{99}}{{99}^{99}}<99$ since $(\frac{(1+n{)}^n}{n^n}\le n)$
$(1+99{)}^{99}<{99}^{99+1}$
$(1+99{)}^{99}<{99}^{100}$
$(100{)}^{99}<{99}^{100}$