Question

solve this in your notebook and post to me​

Answer 1

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

$\sf \large\longmapsto \dfrac{81 \times {3}^{n + 1} - 9 \times {3}^{n} }{81 \times {3}^{n + 2} - 3 \times {3}^{n + 2} }$

=>81 can also be written as 3×3×3×3 or 3⁴

$\sf \large\longmapsto \dfrac{ {3}^{4} \times {3}^{n + 1} - {3}^{2} \times {3}^{n} }{ {3}^{4} \times {3}^{n + 2} - 3 \times {3}^{n + 2} }$

Concept

• If bases are same then their power gets added in multiplication and gets substracted during division.

$\sf \large\boxed{\bold{ \red{ \dfrac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n} }}}$

$\sf \large\boxed{\bold{ \red{ {a}^{m} \times {a}^{n} = {a}^{m + n} }}}$

$\sf \large\longmapsto {3}^{0} \times {3}^{n + 1 - (n + 2)} - {3}^{2 - 1} \times {3}^{n - (n + 2)}$

$\sf \large\boxed{\bold{ \green{ {a}^{0} = 1}}}$

❛Any non-zero value having zero power then it is equals to one ❜

$\sf\large \longmapsto1 \times {3}^{n + 1 - n - 2} - 3 \times {3}^{n - n - 2}$

$\sf\large \longmapsto1 \times {3}^{1 - 2} - 3 \times {3}^{ - 2}$

$\sf \large\longmapsto {3}^{ - 1} - {3}^{ - 2 + 1}$

$\sf \large\longmapsto {3}^{ - 1} - {3}^{ - 1}$

$\sf\large \boxed{\bold{ \blue{ \dfrac{1}{x} = {x}^{ - 1} }}}$

$\sf\large \longmapsto \dfrac{1}{3} - \dfrac{1}{3} = 0$