<s>[INST] Evaluate $\frac{1+3}{3}+\frac{1+3+5}{3^2}+\frac{1+3+5+7}{3^3}+\cdots$ It can be rewritten as
$$S = \frac{2^2}{3}+\frac{3^2}{3^2}+\frac{4^2}{3^3}+\cdots$$
When $k$ approaches infinity, the term $\frac{(k+1)^2}{3^k}$ approaches zero. But, i wonder if it can be used to determine the value of $S$. Any idea?
Note: By using a programming language, i found that the value of $S$ is $3.5$
 [/INST] Let $\dfrac{(k+1)^2}{3^k}=f(k)-f(k-1)$ where $f(n)=\dfrac{an^2+bn+c}{3^n}$
$\implies\dfrac{(k+1)^2}{3^k}=\dfrac{(ak^2+kb+c)-3(a(k-1)^2+b(k-1)+c)}{3^k}$
$\implies k^2+2k+1=-2ak^2+k(b+6a-3b)+c-3(a-b+c)$
Compare the coefficients of $k^2$ to find $$-2a=1$$
and by the coefficients of $k,$ $$6a-2b=2\iff b=?$$
and by the constants, $$1=3b-3a-2c\iff c=?$$
Now use Telescoping series
Of course we need to establish $$\lim_{n\to\infty}f(n)=0$$
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