Q:

The sequence $15,35,80,120,...$ (positive multiples of 5 that are one less than a perfect sq.)...

The sequence $15,35,80,120,...$ consists of positive multiples of 5 that are one less than a perfect square. What are the last 3 digits of the 2017 term?

I know that natural number, when squared, ends in $1,4,6,9$,  and that there is a pattern of two terms ending in 5 and then two terms ending in 0. 
I have managed to figure out that the last digit is 5, but I'm stuck on how to find the other two digits.  

A:

$n^2-1\equiv 0\pmod{5}$ iff $n\equiv 1\pmod{5}$ or $n\equiv 4\pmod{5}$.
Rather than looking at the terms in the form $n^2-1$, let us look at the terms in the form $n$.  These would be $4,6,9,11,14,16,19,21,\dots$.  Once every four terms it increases by ten.  Referring to $4$ as the first element in the sequence, the $2017$'th term of the sequence would then be $4+\frac{2016}{4}\cdot 10=5044$
The $2017$'th term of your original sequence is then $5044^2-1$ which you can either calculate completely, or just calculate the final three digits of while ignoring all of the arithmetic that would affect or carry over to the remaining digits.

 The $2017$'th term in your original sequence is $25441935$

A:

For the odd-numbered terms, $15,80,195,\ldots$, the numbers being squared are $4,9,14,19,\ldots$.  You're looking for the $1009$th term in this sequence (which you then need to square and subtract $1$). The odd-numbered terms, $4,14,24,34,\ldots$ in this sequence are of the form $10n-6$, with $n$ starting at $1$.  You want the $505$th term here, i.e., $5050-6=5044$. So you want the last three digits of $5044^2-1$.  It suffices to compute $44^2-1=1935$ and then drop the lead $1$, to get $935$ as the final three digits.

