Sum binomial coefficients $$\sum_{k=0}^{n} {\frac{k^2+k}{3^{k+2}} {n \choose k}}=?$$
What I've tried:
$$(k^2+k){n \choose k}=k(k+1){\frac{n!}{k!(n-k)!}}$$
$$k^2+k = k^2-k+2k=k(k-1)+2k$$ =>
$$
\begin{align}
(k^2+k){n \choose k} &= k(k-1){\frac{n!}{k!(n-k)!}}+2k{\frac{n!}{k!(n-k)!}}\\&={\frac{n!}{(k-2)!(n-k)!}}+2{\frac{n!}{(k-1)!(n-k)!}}\\&=n(n-1){n-2 \choose k-2}+2n{n-1 \choose k-1}
\end{align}
$$
and
$${\frac{1}{3^{k+2}}}={\frac{1}{9}}({\frac{1}{3}})^k$$
So I have
$$\sum_{k=0}^{n} {\frac{n(n-1){n-2 \choose k-2}+2n{n-1 \choose k-1}}{9*3^k}}$$
And I'm stuck... Can anyone help me?

As in the first hint
$$(\ x(1+x)^n\ )''= \left(\sum_{k=0}^n {n\choose k}x^{k+1}\right)''= \sum_{k=1}^n k(k+1){n\choose k}x^{k-1}= f(x)$$
So if you take $x= \frac{1}{3}$ :
$$\frac{1}{27}f(\frac{1}{3})= \frac{1}{3^3}\sum_{k=1}^n k(k+1){n\choose k}\frac{1}{3^{k-1}}= \sum_{k=1}^n \frac{k(k+1)}{3^{k+2}}{n\choose k}= \sum_{k=0}^n \frac{k(k+1)}{3^{k+2}}{n\choose k}- 0$$
And differentiate :
$$f(x)= (\ (1+x)^n+ nx(1+x)^{n-1}\ )'= 2n(1+x)^{n-1}+ n (n-1)x(1+x)^{n-2}$$
You can compute $f(\frac{1}{3})= 2n(1+\frac{1}{3})^{n-1}+ n (n-1)\frac{1}{3}(1+\frac{1}{3})^{n-2}= 2n(\frac{4}{3})^{n-1}+ n(n-1)\frac{1}{3}(\frac{4}{3})^{n-2}$
