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

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The sequence 15+30+60+120+240+480+960 sums to 1905, and the discussion focuses on determining the $p$th term of the sequence. The quadratic equation derived from the sum is used to find $n$, leading to the formula for the smallest integer greater than $n$. The digits 1-5 repeat in the sequence, allowing for the calculation of the $p$th digit using a modular approach. The conclusion confirms that the $p$th digit for $p=2017$ is 4.
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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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