Question

the value of 2(3power(n+1)+7(3 power (n-1)divided by 3power(n+2)-2(1/3)power(1-n) is​

Answer 1

$\textbf{Given:}$

$\displaystyle\frac{2\;3^{n+1}+7\;3^{n-1}}{3^{n+2}-2(\frac{1}{3})^{1-n}}$

$=\displaystyle\frac{2\;3^n3^1+7\;3^n3^{-1}}{3^n3^2-2\;3^{n-1}}$

$\text{Taking out 3^n common from both numerator and denominator}$

$=\displaystyle\frac{3^n(2\;3^1+7\;3^{-1})}{3^n(3^2-2\;3^{-1})}$

$=\displaystyle\frac{2\;3^1+7\;3^{-1}}{3^2-2\;3^{-1}}$

$=\displaystyle\frac{6+\frac{7}{3}}{9-\frac{2}{3}}$

$=\displaystyle\frac{\frac{18+7}{3}}{\frac{27-2}{3}}$

$=\displaystyle\frac{\frac{25}{3}}{\frac{25}{3}}$

$=\displaystyle\;1$

$\therefore\bf\displaystyle\frac{2\;3^{n+1}+7\;3^{n-1}}{3^{n+2}-2(\frac{1}{3})^{1-n}}=1$

Answer 2

Answer:

Given:

\displaystyle\frac{2\;3^{n+1}+7\;3^{n-1}}{3^{n+2}-2(\frac{1}{3})^{1-n}}

3

n+2

−2(

3

1

)

1−n

23

n+1

+73

n−1

=\displaystyle\frac{2\;3^n3^1+7\;3^n3^{-1}}{3^n3^2-2\;3^{n-1}}=

3

n

3

2

−23

n−1

23

n

3

1

+73

n

3

−1

\text{Taking out $3^n$ common from both numerator and denominator}Taking out 3

n

common from both numerator and denominator

=\displaystyle\frac{3^n(2\;3^1+7\;3^{-1})}{3^n(3^2-2\;3^{-1})}=

3

n

(3

2

−23

−1

)

3

n

(23

1

+73

−1

)

=\displaystyle\frac{2\;3^1+7\;3^{-1}}{3^2-2\;3^{-1}}=

3

2

−23

−1

23

1

+73

−1

=\displaystyle\frac{6+\frac{7}{3}}{9-\frac{2}{3}}=

9−

3

2

6+

3

7

=\displaystyle\frac{\frac{18+7}{3}}{\frac{27-2}{3}}=

3

27−2

3

18+7

=\displaystyle\frac{\frac{25}{3}}{\frac{25}{3}}=

3

25

3

25

=\displaystyle\;1=1

\therefore\bf\displaystyle\frac{2\;3^{n+1}+7\;3^{n-1}}{3^{n+2}-2(\frac{1}{3})^{1-n}}=1∴

3

n+2

−2(

3

1

)

1−n

23

n+1

+73

n−1

=1