$\frac{3^n - 5 \cdot 3^{n-1} + 3^{n-2}}{6^{n+1}} = -\frac{5}{27}$

Question

$$\frac{3^n - 5 \cdot 3^{n-1} + 3^{n-2}}{6^{n+1}} = -\frac{5}{27}$$

I currently don't have any idea about where to start. Can you take a look?

My attempt:

$$\frac{3^n - 5 \cdot 3^{n-1} + 3^{n}\cdot 3}{6^{n+1}} = -\frac{5}{27}$$

$$\frac{3^n (-5 \cdot + 3^{-2})}{6^{n+1}} = -\frac{5}{27}$$

Regards,

Start with $,3^n - 5 \cdot 3^{n-1} + 3^{n-2}=3^{n-2}(3^2 - 5 \cdot 3 + 1),$ and go from there. — dxiv, Mar 19 '18 at 04:59
Are you trying to solve for $n$ or to prove this is an identity? — Ross Millikan, Mar 19 '18 at 05:02
@Displayed I pulled a factor of $3^{n-2}$ out of the numerator, which will cancel out with some part of the denominator after you write that one as $,2^{n+1} \cdot 3^{n+1},$. — dxiv, Mar 19 '18 at 05:10

score 0 · Answer 1 · answered Mar 19 '18 at 05:04

0

Use the substitution $K = 3^{n-2}$ and exponent/power rules.

answered Mar 19 '18 at 05:04

octave

score 0 · Accepted Answer · answered Mar 19 '18 at 05:21

$$\frac{3^n - 5 \cdot 3^{n-1} + 3^{n-2}}{6^{n+1}} = -\frac{5}{27}$$

$$\frac{3^n (1- \frac{5}3 + \frac19)}{6^{n}\cdot 6} = -\frac{5}{27}$$

$$\left(\frac12\right)^n\frac{ (1- \frac{5}3 + \frac19)}{ 6} = -\frac{5}{27}$$

Try to isolate $\left(\frac12\right)^n$ and then take logarithm to complete the question.

2 Answers2