Question

Simplify :
(i)
$\frac{ {81}^{n} . \: 3^{5} - {(3)}^{4n - 1} (243) }{ ({9}^{2n})( {3}^{3}) }$
(ii)
$\frac{4(3)^n}{ {3}^{n + 1} - {3}^{n} }$
​

Answer 1

ANSWER:

(i). 6

(ii). 2

EXPLANATION:

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

$=\frac{( {3}^{4})^{n} . \: 3^{5} - {3}^{4n }. \: {3}^{ - 1} . \: { 3}^{5} }{ ({ {3}^{3}) }^{2n}.( {3}^{3}) }$

$=\frac{{3}^{4n} . \: 3^{5} - \: {3}^{4n }. \: {3}^{4} }{ { {3}^{4n}}. \: {3}^{3}}$

$=\frac{{3}^{4n} . \: 3^{5} }{ { {3}^{4n}}. \: {3}^{3}} - \frac{\: {3}^{4n }. \: {3}^{4}}{\: {3}^{4n }. \: {3}^{3}}$

$= \frac{ {3}^{5}}{ {3}^{3} } - \frac{ {3}^{4} }{ {3}^{3} }$

$= {3}^{5 - 3} - {3}^{4 - 3}$

$= {3}^{2} - 3$

$= 9 - 3$

$= 6$

________________________________________________________________________________________________

$\frac{4(3)^n}{ {3}^{n + 1} - {3}^{n} }$

$= \frac{4(3^n)}{ {3}^{n}(3 - 1) }$

$= \frac{ {4( {3}^{n}) } }{3^{n}(2) }$

$= \frac{4}{2}$

$= 2$

I hope it helps you. Please mark me as BRAINLIEST if you like.