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.
 A: Let $\displaystyle  S=5+11+19+29+41+\cdots+T_{n-1}+T_n$
$\displaystyle S=5+11+19+29+41+\cdots+T_{n-1}+T_n$
$\displaystyle S-S=5+\underbrace{(11-5)+(19-11)+(29-19)+(41-29)+(T_n-T_{n-1})}-T_n$
$\displaystyle\implies T_n-5=6+8+10+12$ upto $n-1$terms  $=(n-1)\{2\cdot6+(n-2)2\}=(n-1)(n+4)$
$\displaystyle T_n=5+n^2+3n-4=n^2+3n+1$
$\displaystyle\implies S_n=\sum_{r=1}^nr^2+3\sum_{r=1}^nr+\sum_{r=1}^n1$
A: If you look at your data using any plotting device, you should notice that the points you give perfectly align along a parabola $$1+3n+n^2$$ Then, the elegant solution given by lab bhattacharjee.
A: The difference table for the sequence is given by
${\color{red}5}\;\;\;\;  11 \;\;\;\; 19  \;\;\;\;29$
$\;\;\;{\color{red}6}\;\;\;\;\;8\;\;\;\;10$
$\;\;\;\;\;\;{\color{red}2}\;\;\;\;\;2$
$\;\;\;\;\;\;\;\;\;0$, 
so $\displaystyle a_n={\color{red}5}\binom{n-1}{0}+{\color{red}6}\binom{n-1}{1}+{\color{red}2}\binom{n-1}{2}$, and therefore
$\displaystyle \sum_{k=1}^{n} a_{k}={\color{red}5}\binom{n}{1}+{\color{red}6}\binom{n}{2}+{\color{red}2}\binom{n}{3}=5n+6\cdot\frac{n(n-1)}{2}+2\cdot\frac{n(n-1)(n-2)}{6}=\frac{n(n+2)(n+4)}{3}.$
A: 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.
A: No one thought of this I guess , but I need further insight into this
$5+11+19+29+41+.....$
$= 5+(10+1+20-1)+(30-1+40+1)+.....(10(n-2)-1+ 10(n-1)+1)$
I assumed for once that n is an odd number , if n is even , it'll leave me with either (10(n-1)-1) or (10(n+1)+1), so let's take n as odd.
We get , 
$5+10+20+30+40+.....+10(n-1)$
$= 5+10{1+2+3+4+.....(n-1)}$
$= 5+10[{(n-1)/2}*(1+n-1)]    [S=(n/2)* (a+l)]$
$= 5+10n(n-1)/2$
$= 5+5n²-5n$
$=5( n²-n-1)$
Now this doesn't even satisfy the equation so correct me where I'm wrong and how this should have been done , by eliminating 1 and turning them into factors of 10
