Question

plz solve........to be brainlist​

Answer 1

Solution :

Now, $\mathrm{\frac{2^{n+3}-2(2^{n})}{2^{2n+2}}}$

$\mathrm{=\frac{2^{3}(2^{n})-2(2^{n})}{2^{2n}\times 2^{2}}}$

$\mathrm{=\frac{(8-2)2^{n}}{4(2^{2n})}}$

$\mathrm{=\frac{6\times 2^{n}}{4\times 2^{2n}}}$

$\mathrm{=\frac{6}{4}\times \frac{2^{n}}{2^{2n}}}$

$\mathrm{=\frac{3}{2}\times 2^{n-2n}}$

$\mathrm{=\frac{3}{2}\times 2^{-n}}$

$\mathrm{=3\times 2^{-n-1}}$

$\to \boxed{\mathrm{\frac{2^{n+3}-2(2^{n})}{2^{2n+2}}=3\times 2^{-n-1}}}$

Laws of Indices :

$\mathrm{1.\:a^{m}\times a^{n}=a^{m+n}}$

$\mathrm{2.\:a^{-m}=\frac{1}{a^{m}}}$

$\mathrm{3.\:(a^{m})^{n}=a^{mn}}$

$\mathrm{4.\:\sqrt[n]{a^{m}}=a^{\frac{m}{n}}}$

Answer 2

Answer :

$\frak{\huge{ \bf{3 \times {2}^{ - n - 1} }}}$

Step By Step Explanation:

$\sf{\frac{ {2}^{n \: + \: 3} \: - \: 2 ({2}^{n}) }{ {2}^{2n \: + \: 2} } }$

$\sf{ \implies \frac{ {2}^{3}(2n ) \: - \:2(2n) }{ {2}^{2n \: + \: 2} } }$

$\sf{ \implies \frac{(8 \: - \: 2) {2}^{n} }{4( {2}^{2n})}}$

$\sf{ \implies \frac{6 \: \times \: {2}^{n} }{4 \: \times \: {2}^{2n} } }$

$\sf{ \implies \frac{ \cancel6}{ \cancel4} \times \frac{ {2}^{n} }{ {2}^{2n} } }$

$\sf{ \implies \frac{3}{2} \times {2}^{n \: - \: 2n} }$

$\sf{ \implies \frac{3}{2} \times {2}^{ - n} }$

$\sf{ \implies 3 \times {2}^{ - n - 1} }$

Laws of Indices :

$\sf{1. \: {a}^{m} \times {a}^{n} = {a}^{m \: + \: n} }$

$\sf{ 2. \: {a}^{m} \: \div \: {a}^{n} } = {a}^{m \: - \: n} \: when \: m > n \\ \: \: \: \: \sf{ {a}^{m} \: \div \: {a}^{n} = \frac{1}{ {a}^{n \: - \: m} } \: when \:m < n}$

$\sf{3. \:( {a}^{m} )^{n}} = {a}^{m \: \times \: n} , (ab)^{m} = {a}^{m} {b}^{m}$

$\sf{ 4. \: {a}^{ - m } = \frac{1}{ {a}^{m}}, \: {a}^{m} = \frac{1}{ {a}^{ - m} } }$

$\sf{5. \: a \degree = 1}$

$\sf{ 6. \: \sqrt[n]{ {a}^{m} } = {a}^{ \frac{m}{n} } }$