<s>[INST] Find 2 sums with the binomial newton Find the sum of:
i)$\displaystyle\sum_{k=0}^{n} k^2$ $\left(\begin{array}{c} n\\k\end{array}\right)$ 
ii) $\displaystyle\sum_{k=1}^{n} \frac{2k+5}{k+1}$$\left(\begin{array}{c} n\\k\end{array}\right)$

Thoughts:
i)(After the Edit)$\displaystyle\sum_{k=0}^{n} k^2$ $\left(\begin{array}{c} n\\k\end{array}\right)$ =
$\displaystyle\sum_{k=0}^{n} {k(k-1)}$ $\left(\begin{array}{c} n\\k\end{array}\right)$ + $\displaystyle\sum_{k=0}^{n}{k}$ $\left(\begin{array}{c} n\\k\end{array}\right)$=
$\displaystyle\sum_{k=0}^{n} {n(k-1)}$ $\left(\begin{array}{c} n-1\\k-1\end{array}\right)$ + $\displaystyle\sum_{k=0}^{n}{n}$ $\left(\begin{array}{c} n-1\\k-1\end{array}\right)$=
$\displaystyle\sum_{k=0}^{n} {n(n-1)}$ $\left(\begin{array}{c} n-2\\k-2\end{array}\right)$ + n$\displaystyle\sum_{k=0}^{n}$ $\left(\begin{array}{c} n-1\\k-1\end{array}\right)$=
$n(n-1)2^{n-2}+n2^{n-1}$=$n2^{n-2}(2+n-1)$=$n(n+1)2^{n-2}$
ii)(After the Edit)$\displaystyle\sum_{k=1}^{n} \frac{2k+5}{k+1}$$\left(\begin{array}{c} n\\k\end{array}\right)$=
$\displaystyle\sum_{k=1}^{n} \frac{(2k+2)+3}{k+1}$$\left(\begin{array}{c} n\\k\end{array}\right)$=
2$\displaystyle\sum_{k=1}^{n}$$\left(\begin{array}{c} n\\k\end{array}\right)$ + 3$\displaystyle\sum_{k=1}^{n} \frac{1}{k+1}$$\left(\begin{array}{c} n\\k\end{array}\right)$=
$2^{n+1}-2$+ $\frac{3}{n+1}$$\displaystyle\sum_{k=1}^{n} \frac{n+1}{k+1}$$\left(\begin{array}{c} n\\k\end{array}\right)$=
$2^{n+1}-2$+ $\frac{3}{n+1}$$\displaystyle\sum_{k=1}^{n}$$\left(\begin{array}{c} n+1\\k+1\end{array}\right)$=
$2^{n+1}-2+\frac{3}{n+1}(2^{n+1}-1)$
 [/INST] Hint:
ii) We can expand as follows
\begin{align*}
\sum_{k=1}^n\frac{2k+5}{k+1}{n\choose k}&=\sum_{k=1}^n\frac{(2k+2)+3}{k+1}{n\choose k}\\[4pt]
&=\sum_{k=1}^n 2{n\choose k}+\frac{3}{n+1}\sum_{k=1}^n\frac{n+1}{k+1}{n\choose k}\\[4pt]
&=2\sum_{k=1}^n{n\choose k}+\frac{3}{n+1}\sum_{k=1}^n{{n+1}\choose {k+1}}
\end{align*}

Notice $$\sum_{k=0}^n{n\choose k}=2^n\qquad\implies\qquad\sum_{k=1}^n{n\choose k}=2^n-1,$$
then,
\begin{align*}
\sum_{k=1}^n\frac{2k+5}{k+1}{n\choose k}&=2\sum_{k=1}^n{n\choose k}+\frac{3}{n+1}\sum_{k=1}^n{{n+1}\choose {k+1}}=2(2^n-1)+\frac{3}{n+1}(2^{n+1}-1)
\end{align*}
Which can be reduced as
$$\sum_{k=1}^n\frac{2k+5}{k+1}{n\choose k}=\frac{2^{n+1}(n+4)-2n-5}{n+1}$$
</s>