How could I find the partial sum of this function? $$\displaystyle\sum_{k = 0}^{n}\dfrac{k^2}{2^k}$$ 
What are the steps to finding the partial sum formula?

Let 
$$S(x)=1^2 x+2^2x^2+3^2x^3+\cdots +n^2x^n.$$ 
Then 
$$\frac{S(x)}{x}=1^2 +2^2 x +3^2 x^2 +\cdots +n^2 x^{n-1}.$$
Note that $\frac{S(x)}{x}$ is the derivative of $T(x)$, where 
$$T(x)=x+2x^2+3x^3+\cdots +nx^n.$$
We have 
$$\frac{T(x)}{x}=1+2x+3x^2+\cdots +nx^{n-1}.$$
Note that $\frac{T(x)}{x}$ is the derivative of $1+x+x^2+x^3+\cdots +x^n$. And this sum is $\frac{1-x^{n+1}}{1-x}$.
Now work backwards. Find the derivative of $\frac{1-x^{n+1}}{1-x}$. Multiply by $x$ to get $T(x)$. Differentiate. That gives us $\frac{S(x)}{x}$. Multiply by $x$ to get $S(x)$. Finally, put $x=\frac{1}{2}$. 
Remark: If calculus is unfamiliar, there are simpler ways! But the calculation above is a useful technique. 
