Question

$4^{-2}+4^{-1}+4^0+4^2$ = ___.

• A. $\frac{277}{16}$
• B. $\frac{277}{18}$
• C. $\frac{257}{16}$
• D. $\frac{271}{18}$

Answer 1

The correct option is A

$\frac{277}{16}$

Note that $a^0=1$ for any non-zero integer $a$.
Also, for integers $a(\ne 0)$ and $m,$
$a^{-m}=\frac{1}{a^m}.$
$⟹4^{-2}=\frac{1}{4^2}=\frac{1}{16}$ and $4^{-1}=\frac{1}{4^1}$
$\therefore 4^{-2}+4^{-1}+4^0+4^2=\frac{1}{16}+\frac{1}{4}+1+16$
$=\frac{1+4+16+256}{16}$
$=\frac{277}{16}$