Question

Evaluate : Please help​

Answer 1

Solution:-

$\rm \to \: \dfrac{ {a}^{2n + 1}a^{(2n + 1)(2n - 1)} }{ {a}^{n(4n - 1)}(a ^{2} )^{2n + 3} }$

Use this identity

$\rm \to \: (x - y)(x + y) = {x}^{2} - {y}^{2}$

Now use this identity

$\rm \to \: \dfrac{ {a}^{2n + 1} {a}^{(2n) {}^{2} - 1} }{ {a}^{n(4n - 1)} ({a}^{2} ) {}^{2n + 3} }$

Use exponential law

$\rm \to \: {a}^{m} \times {a}^{n} = {a}^{m + n}$

$\rm \to \: (a {}^{x} ) {}^{y} = {a}^{xy}$

$\rm \to \: \dfrac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$

Now we get

$\rm \to \: \dfrac{a^{2n + 1}a {}^{4 {n}^{2} - 1} }{ {a}^{4 {n}^{2} - n} {a}^{4n + 6} }$

$\rm \to \: \dfrac{ {a}^{2n + 1 + 4 {n}^{2} - 1 } }{ {a}^{4 {n}^{2} - n + 4n + 6} }$

$\rm \to \: \dfrac{ {a}^{2n + 4 {n}^{2} } }{ {a}^{4 {n}^{2} + 3n + 6} }$

$\rm \to {a}^{2n + 4 {n}^{2} - 4 {n}^{2} - 3n - 6 }$

$\rm \to {a}^{2n + \cancel{4 {n}^{2}} - \cancel {4 {n}^{2} }- 3n - 6 }$

$\rm \to {a}^{ - n - 6}$

$\rm \to \: {a}^{ - (n + 6)}$

$\rm \to \: \dfrac{1}{a {}^{n + 6} }$

Answer

$\rm \to \: \dfrac{1}{a {}^{n + 6} }$