Question

$\frac{(81) {}^{n} . {3}^{5} - {3}^{4n - 4} .243 }{ {9}^{2n} .3} - \frac{4. {3}^{n} }{ {3}^{ n+1 } - {3}^{n} }$
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Answer 1

Answer:

$\boxed{\huge{78}}$

Step-by-step explanation:

$\bold{Given}\longrightarrow \\ \dfrac{ {(81)}^{n} \: {3}^{5} - {3}^{4n - 4} \: 243}{ {9}^{2n} \: 3 } - \dfrac{4 \: ({3}^{n}) }{ {3}^{n + 1} - {3}^{n} }$

$\bold{To \: find}\longrightarrow \\ value \: of \: given \: expression$

$\bold{Concept \: used}\longrightarrow \\ \: \boxed{\huge{ {a}^{m + n} \: = {a}^{m} \: {a}^{n} }} \\ \boxed{\huge{ { ({a}^{m} )}^{n} = {a}^{mn}}}$

$\bold{Solution}\longrightarrow \\ \dfrac{ {(81)}^{n} \: {3}^{5} - {3}^{4n - 4} \: 243}{ {9}^{2n} \: \: 3} - \dfrac{4 \: \: {3}^{n} }{ {3}^{n + 1} \: - {3}^{n} }$

$= \dfrac{ { ({3}^{4} )}^{n} \: {3}^{5} - {3}^{4n - 4} \: {3}^{5} }{ { ({3}^{2}) }^{2n} \: \: 3 } - \dfrac{4 \: {3}^{n} }{ {3}^{n} \: 3 \: - {3}^{n} }$

$= \dfrac{ {3}^{4n} \: {3}^{5 } - {3}^{4n - 4 + 5} }{3 \: {3}^{4n} } - \dfrac{4 \: \: \: ( {3}^{n} )}{ {3}^{n}(3 - 1) }$

$= \dfrac{ {3}^{4n + 5} - {3}^{4n + 1} }{3 \: \: {3}^{4n} } - \dfrac{4}{2}$

$= \dfrac{ {3}^{4n} \: {3}^{5} - {3}^{4n} \: 3 }{3 \: \: {3}^{4n} } - 2$

$= \dfrac{ {3}^{4n} \: ( {3}^{5} - 3) }{3 \: \: {3}^{4n} } - 2$

$= \dfrac{243 - 3}{3} \: \: - \: \: 2$

$= \dfrac{240}{3} \: \: - \: \: 2$

$= 80 \: \: - \: \: 2$

$= 78$