How to calculate the nth term of sequence that increases by n? I have the following recursive formula the for a sequence:
\begin{cases}V_{1} = 1\\V_{n} = V_{n-1} + n & n > 1\end{cases}
This sequence increases by $n$ for each term increase. Now I need to find if $3003$ is a value of this sequence. I managed to solve the problem by finding an explicit formula for the sequence and solved in terms of $n$. Like this:
\begin{align}
3003 = \dfrac {\left( n-1\right) n} {2}+n 
\end{align}
which led to 
\begin{align}
n = 77 &\vee n = -78
\end{align}
So since there's no negative terms in a sequence I concluded that when $n=77$,  $3003$ is the value of the sequence. 
What I would like to know is if there is some other simpler/direct way to solve this.
I realized that the value of the nth term is the sum of all terms below $n$, including it $(n, n-1, n-2,\dots,0)$ and came up with that explicit formula but it isn't obvious. 

Is it obvious now?
$$\begin{align}V_n&=n+V_{n-1}\\&=n+(n-1)+V_{n-2}\\&=n+(n-1)+(n-2)+V_{n-3}\\&=\ \vdots\\&=V_1+2+3+\dots+n\\&=1+2+3+\dots+n\end{align}$$

To avoid the explicit formula for $V_n$, one could use calculus, which quickly shows, using integrals, that
$$\frac{n^2}2<V_n<\frac{(n+1)^2}2$$

Once you've done that and figured out that $V_n=\frac{n(n+1)}2$, it follows that we want to solve
$$3003=\frac{n(n+1)}2$$
which may be 'solved' by noticing that
$$\frac{n^2}2<\frac{n(n+1)}2<\frac{(n+1)^2}2$$
which gives
$$76.498=\sqrt{6006}-1<n<\sqrt{6006}=77.498$$
If $n$ is to be a whole number, then
$$n=77$$
without ever having to solve hard quadratics.
