Question

3.9^n+1 +9.3^2n-1÷9.3^2n-3.9^n-1

Answer 1

you mean, $\frac{3.9^{(n+1)}+9.3^{(2n-1)}}{9.3^{(2n)}-3.9^{(n-1)}}$

we can write 9 = 3²

so, 9^(n + 1) = {3²}^(n + 1) = 3^{2(n + 1)}

= 3^(2n + 2)

similarly, 9^(n - 1) = 3^(2n - 2)

now, {3.3^(2n + 2) + 9.3^(2n - 1)}/{9.3^(2n)-3.3^(2n - 2)}

= {3¹.3^(2n + 2) + 3².3^(2n - 1)}/{3².3^(2n) - 3¹.3^(2n - 2)}

we know, a^m . a^n = a^(m + n)

so, 3¹.3^(2n + 2) = 3^(2n + 2 + 1) = 3^(2n + 3)

similarly, 3².3^(2n - 1) = 3^(2n + 1)

3². 3^(2n ) = 3^(2n + 2)

3¹.3^(2n - 2) = 3^(2n - 1)

now, {3¹.3^(2n + 2) + 3².3^(2n - 1)}/{3².3^(2n) - 3¹.3^(2n - 2)} = {3^(2n + 3) + 3^(2n + 1)}/{3^(2n + 2) - 3^(2n - 1)}

= {3^(2n + 1)(3² + 1)}/{3^(2n - 1)(3³ - 1)}

= {3^(2n + 1)/3^(2n - 1)}[(3² + 1)/(3³ - 1)]

we know, x^m/x^n = x^(m - n)

so, 3^(2n + 1)/3^(2n - 1) = 3^(2n + 1 - 2n + 1) = 3² = 9

so, {3^(2n + 1)/3^(2n - 1)}[(3² + 1)/(3³ - 1)] = 9 × 10/26

= 90/26