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༒ Bʀᴀɪɴʟʏ Sᴛᴀʀs

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$\dashrightarrow \ \sf \dfrac{1}{n!} \ - \ \dfrac{1}{(n - 1)!} \ - \ \dfrac{1}{(n - 2)!}.$


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Answer 1

Step-by-step explanation:

$\\ \\ \\ \\ \\ \sf\underline{Laws \: of \: exponents \: (Integer \: exponents)} \: \\ \\ \\ \sf \: a\neq0 \: and \: b\neq0 and integers x \: and y , \\ \\\bigstar \sf a^{0}=1 and a^{-n}=\dfrac{1}{a^{n}} \\ \\ \sf \bullet a^{x} \times a^{y}=a^{x+y} \\ \\\bullet \sf {a}^{x} \div {a}^{y}={a}^{x-y} \\ \\\bullet \sf({a}^{x})^{y} =a^{xy} \\ \\\bullet \sf (ab)^{x}=a^{x}b^{x}\\ \\ \\ \\ \\ \sf \: For \: \: \neq0 and \: b \: \neq0 and \: rationals \: r \: \: and \: \: s \\ \\ \bigstar \sf \: a^{\frac{1}{n}}=\sqrt[n]{a} \: \: and \: \: a^{\frac{m}{n}}=\sqrt[n]{a^{m}} \\ \\\bullet \sf \: a^{r} \times a^{s}=a^{r+s} \\ \\\bullet \sf \: {a}^{r} \div {a}^{s}={a}^{r-s} \\ \\ \sf \: \bullet ({a}^{r})^{s} =a^{rs} \\ \\\bullet \sf \: (ab)^{r}=a^{r}b^{r}\\ \\ \\ \\ \sf \: \underline{Laws \: of \: exponents \: (Real \: exponents)} \\ \ \sf \: For \: a\neq0 and b\neq0 and reals x \: and \: y \\ \\ \sf \: \bullet a^{x} \times a^{y}=a^{x+y} \\ \\\bullet \sf \: {a}^{x} \div {a}^{y}={a}^{x-y} \\ \\\bullet \sf \: ({a}^{x})^{y} =a^{xy} \\ \\\bullet \sf \: (ab)^{x}=a^{x}b^{x}$

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