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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.

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    See [this](http://math.stackexchange.com/questions/1417579/largest-triangular-number-less-than-a-given-natural-number) and [this](http://math.stackexchange.com/questions/698961/finding-the-triangular-root-of-a-number) and [this](http://math.stackexchange.com/questions/455511/formula-for-the-nth-term-of-1-2-2-3-3-3-4-4-4-4-5) on how to check whether a number is triangular – 2017-02-07

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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

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

If $n$ is to be a whole number, then

$$n=77$$

without ever having to solve hard quadratics.

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    Thank you for your response and for pointing me to the prove of that formula. It really makes sense now. But my doubt still remains. It is possible to solve it without knowing or coming up to that formula? – 2017-02-07
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    @AntónioQuadrado If we allow calculus? – 2017-02-07
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    I'm not quite there yet... Isn't possible without it? – 2017-02-07
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    I'll think it over to see if I have some good ideas. – 2017-02-07
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    @AntónioQuadrado The calculus may now be fulfilled (see [here](https://www.desmos.com/calculator/uivebdmjok). The area of the staircase is $V_n$, and the areas of each triangle is easy to do. – 2017-02-07
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    I'm not sure I follow.. Vn wouldn't be a uniform staircase like that but one where each step would have the height of the last one plus the number of the step... Am I wrong? – 2017-02-07
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    @AntónioQuadrado You are not, but you misread my statement. I said $V_n$ was the **area** under the staircase. I then asked you to compare it to the areas of the triangles. – 2017-02-07
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53251/discussion-between-antonio-quadrado-and-simply-beautiful-art). – 2017-02-08