Question

prove 3^n+4 -9.3^n+1/9.3^n+1 = 2

Answer 1

Question :–

$\\ \: \: { \bold { prove \: \: that \: : \: \dfrac{ {3}^{n + 4} - 9. {3}^{n + 1} }{ 9.{3}^{n + 1} } = 2 }}$

ANSWER :–

TO PROVE :–

$\\ \: \implies \: { \bold { \dfrac{ {3}^{n + 4} - 9. {3}^{n + 1} }{ 9.{3}^{n + 1} } = 2 }}$

SOLUTION :–

• Let's take L.H.S. –

$\\ \: \: { \bold { = \dfrac{ {3}^{n + 4} - 9. {3}^{n + 1} }{ 9.{3}^{n + 1} } }}$

• We know that –

$\\ \: \longrightarrow \large \: { \bold { {a}^{b + c} = {a}^{b} . {a}^{c} }}$

• So that –

$\\ \: \: { \bold { = \dfrac{ {3}^{n } . {3}^{4} - 9. 3.{3}^{n } }{ 9.{3}^{n } .3} }}$

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

$\\ \: \: { \bold { = \dfrac{ {3}^{n } . {3}^{4} - { 3}^{3} .{3}^{n } }{ {3}^{n } .{3}^{3}} }}$

$\\ \: \: { \bold { = \dfrac{ \cancel{ {3}^{n } } ( {3}^{4} - { 3}^{3} )}{ \cancel{ {3}^{n }} .{3}^{3}} }}$

$\\ \: \: { \bold { = \dfrac{ ( {3}^{4} - { 3}^{3} )}{{3}^{3}} }}$

$\\ \: \: { \bold { = \dfrac{ 81- 27}{27} }}$

$\\ \: \: { \bold { = \cancel\dfrac{ 54}{27} }}$

$\\ \: \: { \bold { = 2}}$

$\\ \: \: { \bold { = R.H.S. \: \: \: \: \: \: \: \: (Hence \: \: proved)}}$

Answer 2

To Prove :-

$\sf{\implies \frac{ {(3)}^{n + 4} - {(9.3)}^{n + 1} }{(9.3)^{n + 1} } = 2 }$

Solution :-

According to the law of exponent .

$\sf{\implies {a}^{m + n} = {a}^{m} . {a}^{n} }$

$\sf{ \implies \: ({a}^{m} )^{n} = {a}^{mn}}$

Taking LHS :-

$\sf{\implies \frac{ {(3)}^{n + 4} - {(9.3)}^{n + 1} }{(9.3)^{n + 1} } }$

$\sf{\implies \frac{ {(3)}^{n} . {(3)}^{4} - {(3.3.3)}^{n+1} }{(9.3)^{n}.{(9.3)}^{1} } }$

$\sf{\implies \frac{ {(3)}^{n} . {(3)}^{4} - {3}^{2}. 3. {3}^{n} }{(9.3)^{n}.{(9.3)}^{1} } }$

→ 9 × 3 = 3³

$\sf{\implies \: \frac{( {3}^{n}. {3}^{4} )- ( {3}^{3}. {3}^{n} ) }{ {3}^{3}. {3}^{n} } }$

Taking 3^n common in the numerator .

$\sf{ \implies \: \frac{{3}^{n}( {3}^{4} - ( {3}^{3}. ) }{ {3}^{3}. {3}^{n} } }$

$\sf{ \implies \: \frac{\cancel{{3}^{n}}( {3}^{4} - ( {3}^{3}. ) }{ {3}^{3}. \cancel{ {3}^{n}} } }$

$\sf{ \implies \: \frac{81 - 27}{27} = \frac{54}{27} }$

→ 2 . = RHS = LHS

Hence proved