Dtyjlui

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?

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.

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 $kgeq 1$. Then
$$sumlimits_{k=1}^{10} (1.5k^2-0.5k+2)=1.5sumlimits_{k=1}^{10} k^2-0.5sumlimits_{k=1}^{10} k+2sumlimits_{k=1}^{10}1$$ which you can evaluate using the well-known sum formulas.

$x=1 phantom{n} 2 phantom{n} 3 phantom{n} 4$

$y=3 phantom{n} 7 phantom{n} 14 phantom{n} 24$

$y=m_2(x-x_1)(x-x_2)+m_1(x-x_1)+y_1$

$m_1$ 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

$m_2=((y_3-y_2)-(y_2-y_1))/(x_3-x_1)$

$m_2=((14-7)-(7-3))/(3-1)$

$m_2=3/2$

$y=3/2*(x-2)(x-1)+2*(x-1)+3$

$y=1.5x^2-0.5x+4$

Take a difference of the difference, and you get a constant sequence.
This is a quadratic polynomial, not a linear one...

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+2sum_{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}