Summing $3+7+14+24+37...$ up to $10$ terms 
What is $3+7+14+24+37...$ up to $10$ terms? 

I only see that difference between two consecutive terms is in AP ie $7-3=4,14-7=7,24-14=10$ and so on. Any ideas on how to do it?
 A: The sequence and its forward differences:
$$a_n : 3,7,14,24,37,..$$
$$\Delta a_n : 4,7,10,13,..$$
$$\Delta^2 a_n : 3,3,3,...$$
$$\Delta^3 a_n : 0,0,..$$
$$\Delta^4 a_n : 0,..$$
Now we make a reasonable guess that $\Delta^4 a_n=0$ i.e. we guess $\Delta^4 a_n$ is a sequence with all terms as $0$ to get $\Delta^k a_n=0$  when $k \geq 4$ and is an integer:
$$\Delta^5 a_n=0 : 0,..$$
$$\Delta^6 a_n=0 : 0,...$$
..
If we take $a_0$ to be the first term of the sequence than looking at the first term of each of the above sequences $3,4,3,0,0,0,..$  we have:
$$a_n=3{n \choose 0}+4{n \choose 1}+3{n \choose 2}+0{n \choose 3}+0{n \choose 4}+0{n \choose 5}+\cdots$$
$$=3{n \choose 0}+ 4{n \choose 1}+3{n \choose 2}$$
Where now you can apply the summation formula $\sum_{n=0}^{x} {n \choose k}={x+1 \choose k+1}$. You can shift this to the right one if you want $a_1$ to represent the first term instead of $a_0$. Either way, you will get that the sum you want is equal to :

$$3{10 \choose 1}+4{10 \choose 2}+3{10 \choose 3}=570$$

Because there are $10$ terms $a_0,..a_9$ and using the given formula that is what we get.
If you want a proof of what I hope you are observing with the differences let me know. The summation formula was grabbed from Wikipedia.
A: Since the second differences are constant, the terms come from a quadratic polynomial. One can check that the terms of the series come from the polynomial $1.5k^2-0.5k+2$ where $k\geq 1$. Then 
$$\sum\limits_{k=1}^{10} (1.5k^2-0.5k+2)=1.5\sum\limits_{k=1}^{10} k^2-0.5\sum\limits_{k=1}^{10} k+2\sum\limits_{k=1}^{10}1$$ which you can evaluate using the well-known sum formulas.
A: $x=1 \phantom{n} 2 \phantom{n}  3 \phantom{n}  4$
$y=3 \phantom{n}  7 \phantom{n} 14 \phantom{n} 24$
$y=m_2(x-x_1)(x-x_2)+m_1(x-x_1)+y_1$
$m_1$ is slope 
$m_1=(y_2-y_1)/(x_2-x_1)$
$m_1=(7-3)/(2-1)$
$m_1=4$
$m_2$ changes in slope 
$m_2=((y_3-y_2)-(y_2-y_1))/(x_3-x_1)$
$m_2=((14-7)-(7-3))/(3-1)$
$m_2=3/2$
$y=3/2*(x-2)(x-1)+2*(x-1)+3$ 
$y=1.5x^2-0.5x+4$
A: Take a difference of the difference, and you get a constant sequence.
This is a quadratic polynomial, not a linear one...
A: This answer seeks to expand on the "one can check that" comment in Foobaz's answer. First, let $A=\{3,7,14,24,37,\ldots\}$. The pattern for the difference, $d_n$, between terms $n$ and $n+1$ is clear enough:
$$
d_n=4+(n-1)3=1+3n.
$$
Hence,
$$
a_{n+1}=\frac{n}{2}(5+3n)+3
$$
or, more usefully,
\begin{align}
a_n&=\frac{n-1}{2}(2+3n)+3\\[1em]
&=(n-1)+\frac{3n^2-3n}{2}+3\\[1em]
&= \frac{2n-2+3n^2-3n+6}{2}\\[1em]
&= \frac{3n^2-n+4}{2}.
\end{align}
Thus, we have the following:
\begin{align}
S_{10}&= \sum_{i=1}^{10}\frac{3i^2-i+4}{2}\\[1em]
&= \frac{3}{2}\sum_{i=1}^{10}i^2-\frac{1}{2}\sum_{i=1}^{10}i+2\sum_{i=1}^{10}1\\[1em]
&= \frac{3}{2}\left[\frac{10(10+1)(2(10)+1)}{6}\right]-\frac{1}{2}\left[\frac{10(10+1)}{2}\right]+2(10)\\[1em]
&= 570.
\end{align}
