How to calculate the nth term and the sum in this series if the common difference between them isn't explicit? I have this series comprised of
$$1,2,5,10,17,26,...$$,
and so on. So far i have found that they add up in intervals being odd numbers. But I don't know how to find, let's say the 16th term and the sum up to that term. What should I do?
Edit:
Although it is somewhat possible to calculate the sum up to the 16th term in the above equation, what if the series is of higher order like this?.
$$6,7,14,27,46,...$$
Is there a shortcut to calculate the sum up to the 16th term? Can this be done by hand or would be necessary to use software like Maple?.

As Guarang Tandon says. You can create a table of differences between consecutive terms and then repeat that process on the list you just created.
\begin{array}{|l|c|ccccccc|}
\hline
\text{index} & n & 0 & 1 & 2 & 3 & 4 & 5 & \dots\\
\hline
\text{sequence} & f_n & 1 & 2 & 5 & 10 & 17 & 26 & \dots \\
\text{first differences} & \Delta f_n && 1 & 3  & 5 & 7 & 9 & \dots \\
\text{second differences} & \Delta^2 f_n  &&& 2 & 2 & 2 & 2 & \dots \\
\hline
\end{array}
If you are lucky, at some point, all of the differences will be constant. (Of course this is an unverifiable claim. All we know for sure is that the first four second differences are constant.  We can only assume that this fortuitous pattern continues.)
It can be proved that, if $\Delta^d f_n$ is constant, then then $f_n$ is a $d$th degree polynomial in $n$.
If the second differences are constant, then the first differences must be an arithmetic sequence. In this case, we see that $\Delta f_n = 2n+1$. It follows that $$f_{n+1} = f_n + 2n+1 \tag 1$$.
If the first differences are an arithmetic sequence, then $f_n = an^2 + bn + c$ for some real numbers $a, b,$ and $c$.
We can now use modified form of mathematical induction to find the values of $a,b, $ and $c$.
Since $f_0=1$, then $1 = a\cdot0^2 + b\cdot 0 + c = c$. So $f_n=an^2+bn+1$.
First we note that $$f_{n+1} = a(n+1)^2 + b(n+1) + 1 = f_n + 2an + (a+b).$$
Comparing that to $f_{n+1}= f_n+ 2n+1$, we see that $2a=2$ and $a+b=1$. Hence $a=1$ and $b=0$.
We conclude that $f_n=n^2+1$.
We know that $\sum_{k=0}^n k^2=\frac 16n(n+1)(2n+1)$.
Then 
\begin{align}
   \sum_{k=0}^n f_k
   &= \sum_{k=0}^n (k^2+1) \\
   &= \sum_{k=0}^n k^2 + \sum_{k=0}^n 1 \\
   &= \frac 16n(n+1)(2n+1) + (n+1) \\
   &= \frac 16(n+1)(2n^2+n) + \frac 16(n+1)6 \\
   &= \frac 16(n+1)(2n^2+n+6) \\
\end{align}
Check:
$\qquad \sum_{k=0}^5 f_k = 1+2+5+10+17+26=61$
$\qquad \left. \dfrac 16(n+1)(2n^2+n+6)\right|_{n=5}=\frac 16(6)(61)=61$
So, since we started counting at $n=0$,  the $16$th term is $f_{15} = 15^2+1=226$
and the sum of the first $16$ terms is
$\sum_{k=0}^{15} f_k = \left. \dfrac 16(n+1)(2n^2+n+6)\right|_{n=15}=1256$.
