Finding the sum of a series till $n$ terms Series:
5, 11, 19, 29, 41
Find the sum of the series up to $n$ terms.
Well the method that comes to my mind is to find the nth term of the sequence, and then find their summation. I use the basic formulas, such as sum of series $n^2$, $n^3$, etc.
My question is whether their is a shorter method to achieving this, or maybe just an alternate method.

For a slightly different way of getting the pattern, note that the sequence of first differences is $6,8,10,12,\ldots$ and thus the second differences are constant. So the sequence is quadratic i.e. $a_n=An^2+Bn+C$. Then \begin{align}
a_1&=A+B+C=5,\\\\
a_2-a_1&=3A+B=6,\\\\
(a_{n+2}-a_{n+1})-(a_{n+1}-a_n)&=2A=2,
\end{align} which we may backsolve to obtain $(A,B,C)=(1,3,1)\implies a_n=n^2+3n+1$.
As an alternative to summing the series with standard formulae, start by considering the ordinary generating function $A(x)=\sum_{n=1}^\infty a_n x^n$. Since $a_{n+2}-2a_{n+1}+a_n=2$, we may write 
\begin{align}
\sum_{n=1}^\infty a_{n+2} x^{n+2}&=A(x)-(a_2 x^2+a_1 x)=A(x)-(11x^2+5x) ,\\
&=2 \sum_{n=1}^\infty x^{n+2}+2\sum_{n=1}^\infty a_{n+1} x^{n+2}-\sum_{n=1}^\infty a_n x^{n+2}\\
&=\frac{2x^3}{1-x}+2x\cdot\left(A(x)-5x\right)-x^2 A(x)
\end{align}
from which we deduce 
\begin{align}
A(x)
&=\frac{1}{x^2-2x+1}\left(11x^2+5x+\frac{2x^3}{1-x}-10x^2\right)\\
&=\frac{x^3-4x^2+5x}{(1-x)^3}\\
&=5x+11x^2+19x^3+29x^4+41x^5+\ldots.
\end{align}
WolframAlpha bears out the series expansion in the final line.
To use this to obtain the partial sums $S_n=\sum_{k=1}^n a_n$, observe that 
$$\sum_{n=1} S_n x^n =\sum_{n=1}^\infty\sum_{k=1}^n a_n x^n=\sum_{k=1}^\infty\sum_{n=k}^\infty a_k x^n=\sum_{k=1} \frac{a_k x^k}{1-x}=\frac{A(x)}{1-x}$$ where in the second line we have reversed the order of summation, allowing us to sum the geometric series and recognize $A(x)$. So $S_n$ is given by the $n$-th coefficient of $$\frac{A(x)}{1-x}=\frac{x^3-4x^2+5x}{(1-x)^4}=\frac{2}{(1-x)^4}-\frac{1}{(1-x)^2}-\frac{1}{1-x}.$$ The partial fractions can be expanded by the (negative) binomial series, yielding 
\begin{align}
S_n &= 2\binom{n+3}{n}-\binom{n+1}{n}-\binom{n}{n}\\
&=\frac{1}{3}(n+3)(n+2)(n+1)-(n+1)-1\\
&=\frac{1}{3}n(n+2)(n+4)
\end{align}
as the formula for the $n$-th partial sum. As a check, note that this begins as $5,16,35,64,\ldots$ which are indeed the first few partial sums.
Now, this was a fairly algebra-heavy route. But I find it has the advantage of being systematic and rather direct; it's also rewards your knowledge of Taylor series over clever guessing. For instance, note that the observation that $A(x)/(1-x)$ gives the partial sums of the coefficients of $A(x)$ didn't depend on on what $A(x)$ was. So if we're willing to do the work of getting $A(x)$ for a given sequence, then all that remains is to expand $A(x)/(1-x)$ appropriately.
