Find the value of $3+7+12+18+25+\ldots=$ Now, this may be a very easy problem but I came across this in an examination and I could not solve it.
Find the value of
$$3+7+12+18+25+\ldots=$$
Now here is my try
$$3+7+12+18+25+\ldots=\\3+(3+4)+(3+4+5)+(3+4+5+6)+(3+4+5+6+7)+\ldots=\\3n+4(n-1)+5(n-2)+\ldots$$
After that, I could not proceed.
 A: The above sequence is given by $a_i = \frac{i(i+5)}{2}$. The finite sum is given by
\begin{align}
S_n & = \sum_{i=1}^n a_i \\
& = \frac{1}{2} \sum_{i=1}^n i^2 + \frac{5}{2} \sum_{i=1}^n i \\
& = \frac{n(n+1)(2n+1)}{12} + \frac{5n(n+1)}{4} \\
& = \frac{n(n+1)(n+8)}{6}
\end{align}
EDIT: (In response to comments) To a certain point, finding the formula for $a_i$ is trial and error. However, it is not difficult to note here that the difference of the difference of terms is always 1 (i.e. $\Delta^2 a_i = 1$). This implies that the dependence is quadratic. Using $a_i = A i^2 + B i + C$ we can determine the formula with three terms.
A: Although I prefer Gregory's answer that computes this directly, here is another approach:
Notice:
$s_1 = 3$;
$s_2 = 10$;
$s_3 = 22$;
$s_4 = 40$;
$s_5 = 65$
Let $s_n = an^3 + bn^2 + cn + d$
Now, solve the system of equations given by 
$s_1 = 3$;
$s_2 = 10$;
$s_3 = 22$;
$s_4 = 40$;
to find that:
$a = 1/6, b = 3/2, c = 4/3, d = 0$, hence:
$s_n = 1/6n^3 + 3/2n^2 + 4/3n$ 
which yields the same as Gregory's answer.
A: If you know the value of $n$ then $$3+7+12+18+25+\ldots=\\3n-3n+3+7+12+18+25+\ldots$$ As $3=1+2$ we can write as  $$\\1+2+3+(1+2+3+4)+(1+2+3+4+5)+(1+2+3+4+5+6)+\ldots-3n=\\\left(\sum_{n=3}^n{\frac{n(n+1)}{2}}\right)-3n$$ 
A: I'm adding another answer because people ask how to find $a_n$ without trial and error.
We note that:
$a_2 - a_1 = 4; a_3 - a_2 = 5; a_4 - a_3 = 6$, 
which leads us to conclude that $a_n$ is given by the recurrence relation:
$a_{n+1} - a_n = n+3$
Let's start by solving the homogeneous equation:
$a_{n+1} - a_n = 0$
The associated polynomial is $P(r) =r - 1$. The root of this polynomial is $r = 1$. Therefore, a solution to the homogeneous equation is $a_n^{h} = 1^n = 1$.
Now, we want to find a particular solution, so let's try $a_n^{p} = An + B$.
Then:
$A(n+1) + B - An  - B= n+3$. This does not work.
Let's try $a_n^{p} = An^2 + Bn + C$
Thus:
$A(n+1)^2 + B(n+1) + C - An^2 - Bn - C = n +3$
from which follows:
$2A = 1, A + B = 3, C \in \mathbb{R}$. 
Hence:
$A = 1/2;
B = 5/2;
C \in \mathbb{R}$
Therefore, $a_n^p = 1/2n^2 + 5/2n + C$
and 
$a_n = a_h + a_p = 1 + 1/2n^2 + 5/2n + C = 1/2n^2 + 5/2n + D$
Because $a_1 = 3$, it follows that $D = 0$. 
We conclude that $a_n = 1/2n^2 + 5/2n \quad \triangle$
A: This answer is unique in that it does not use trial and error to find out the expression for the $n$-th term. 
Our sum is given by :
$$
S=3+7+12+18+...+T_n
\\
S=\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;3+7+12+18+...+T_n-1+T_n
$$
Subtracting these two sums :
$$
0=3+4+...+(n+2)-T_n
\\
T_n=3+4+...+(n+2)
\\
T_n=\frac{(n+2)(n+3)}{2}-3
=\frac{n(n+5)}{2}
$$
Now, you can use Gregory's answer to figure the sum out. 
