Note that a^0=1 for any non zero integer a. Also, for integers a (≠ 0) and m, a^ m=1/a^m. 4^ 2=1/4^2=1/16 and 4^ 1=1/4^1 nbsp;∴ 4^ 2+ 4^ 1 nbsp;+ 4^0 nbs ...

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Standard VII

Mathematics

Quotient Law

4-2+4-1+40+42...

Question

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

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Solution

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

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4−2+4−1+40+42 = ___.

Note that a0=1 for any non-zero integer a. Also, for integers a (≠0) and m, a−m=1am. ⟹4−2=142=116 and 4−1=141 ∴4−2+4−1+40+42=116+14+1+16 =1+4+16+25616 =27716

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