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

[Ilikebugs]

View attachment 6431

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

 

[MarkFL]

Let's find the $p$th term.

I think what I would do is set:

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

Now solve for $n$...

\(\displaystyle n(n+1)=2p\)

\(\displaystyle n^2+n-2p=0\)

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

\(\displaystyle n=\frac{-1+\sqrt{8p+1}}{2}\)

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

\(\displaystyle \left\lceil \frac{-1+\sqrt{8p+1}}{2}\right\rceil\)

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

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

And we find:

\(\displaystyle D(2017)=4\quad\checkmark\)