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.

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 $a_n = 1/2n^2 + 5/2n \quad \triangle$
