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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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António Quadrado
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1 Answers1

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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

$$\frac{n^2}2<V_n<\frac{(n+1)^2}2$$

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.

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