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.

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. 
