Answer Sequence: 15+30+60+120+240+480+960=1905

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The discussion centers on the mathematical sequence defined by the sum 15+30+60+120+240+480+960, which totals 1905. The participants derive the $p$th term using the formula $$\frac{n(n+1)}{2}=p$$ and solve for $n$ through the quadratic equation $$n^2+n-2p=0$$. The final expression for the $p$th digit is given by $$D(p)=\left\lceil \frac{-1+\sqrt{8p+1}}{2}\right\rceil\mod5$$, confirming that $$D(2017)=4$$ is accurate.

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I did 15+30+60+120+240+480+960 to get 1905, then continued the sequence to get 4. Is this right?
 

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Let's find the $p$th term.

I think what I would do is set:

$$\frac{n(n+1)}{2}=p$$

Now solve for $n$...

$$n(n+1)=2p$$

$$n^2+n-2p=0$$

Using the quadratic formula, and discarding the negative root, we find:

$$n=\frac{-1+\sqrt{8p+1}}{2}$$

We are interested in the smallest integer greater than $n$, so we use

$$\left\lceil \frac{-1+\sqrt{8p+1}}{2}\right\rceil$$

Since the digits 1-5 repeat, then the $p$th digit $D$ is:

$$D(p)=\left\lceil \frac{-1+\sqrt{8p+1}}{2}\right\rceil\mod5$$

And we find:

$$D(2017)=4\quad\checkmark$$
 

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