# Find formula for ${-\frac{2\ln3}{3} , \frac{4\ln4}{9} ,\frac{6\ln5}{27},\frac{8\ln6}{81} ,…}$

The question is to find a general formula for the nth term, $a_n$, of the sequence: $${-\frac{2\ln3}{3} , \frac{4\ln4}{9} ,\frac{6\ln5}{27}, \frac{8\ln6}{81} ,\frac{10\ln7}{243},...}$$

Here is what I got, but when I plugged in a few terms, it did not work out right. I got the formula to be: $$a_n = (-1)^n \frac{2\ln(2+1)\cdot 2^{(n-1)}} {3^n}$$

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One quick comment, my question is about to find the formula of a sequence, not the sum of the series – Jaden Q Oct 30 '12 at 5:09

Each term has a form like $\pm \frac{x\ln y}{z}$.

The sign alternates so it is $(-1)^n$.

$x$ is $2, 4, 6, 8, \dots$, so $x$ is $2n$.

$y$ is $3,4,5,\dots$ so $y$ is $n+2$.

$z$ is $3,9,27, 81, \dots$ so $z$ is $3^n$.

Combining all contributions, we get $a_n = (-1)^n \frac{2n \ln(n+2)}{3^n}$.

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thanks! it totally makes more sense to me – Jaden Q Oct 30 '12 at 5:13

Sign changes in alternative terms,so $(-1)^n$

$2,4,6,8\cdots \implies n$ th term is $2n$

If the question meant, $$\frac{\log 3}{3}, \frac{\log 4}{9}, \frac{\log 5}{27},\cdots$$ its $n$th term is $$\frac{\log (n+2)}{3^n}$$

If the question meant, $$\log \frac{3}{3}, \log \frac{4}{9}, \log \frac{5}{27},\cdots$$ its $n$th term is $$\log \frac{(n+2)}{3^n}$$

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The answer $a_n = (-1)^n \frac{2n \ln(n+2)}{3^n}$ is true if the sequence was:

${-\frac{2\ln3}{3} , \frac{4\ln4}{9} ,-\frac{6\ln5}{27}, \frac{8\ln6}{81} ,-\frac{10\ln7}{243},...}$

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