For each positive integer number $n$ is denoted by $p (n)$ the number of perfect squares nonzero, at most equal to $n$ and $S_n = p (1)+ p (2)+ p (3)\cdots+ p (n)$.
a) Calculate $S_{17}$
b) Determine the lowest positive integer number $n$ for which $S_n > 2017$
If you can, please give me some advice because I do not know how to determine the general term. Thanks!
$p(n) = \lfloor \sqrt{n} \rfloor$
So, $p(1)=1, p(2)=1, p(3)=1, p(4)=2, p(5)=2, p(6)=2, p(7)=2, \ldots$.
Let's look at the pattern here: $S_1 = 1$, $S_2 = 2$, $S_3 = 3$, $S_4 = 5$, $S_5 = 7$, $S_6 = 9$, $S_7 = 11$, $S_8 = 13$, $S_9 = 16$, $S_{10} = 19$, $S_{11} = 22$, $S_{12} = 25$, $S_{13} = 28$, $S_{14} = 31$, $S_{15} = 34$, $S_{16} = 38$, $S_{17} = 42$
Essentially, $S_n = \displaystyle \sum_{k=1}^{\lfloor \sqrt{n} \rfloor}(n-k^2+1) = n-\dfrac{1}{6}\left(\lfloor \sqrt{n} \rfloor -1 \right) \left( 2\lfloor \sqrt{n} \rfloor^2+5\lfloor \sqrt{n} \rfloor - 6n \right)$
So, $S_{17} = 42$.
I leave part (b) to you.