2 problems related to the number 2015 
*

*Let $p=\underbrace{11\cdots1}_\text{2015}\underbrace{22\cdots2}_\text{2015}$. Find $n$, where $n(n+1) = p$

*Prove that $\frac{1}{2^2} + \frac{1}{3^2} + \cdots +  \frac{1}{2015^2} < \frac{2014}{2015}$
For 1, I tried dividing in various ways until I got a simpler expression, but no result. For 2, I know that $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$, but the proofs i found are over the elementary level the problem is aimed for. I also proved that it's smaller than $\frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{1024^2} < 1$, but that obviously is missing $\frac{1}{2015}$
 A: $1)$ $n=\frac{10^{2015}-1}{3}=3...3$ with $2015$ $3's$ :
$$\frac{10^{2015}-1}{3}\times \frac{10^{2015}+2}{3}=\frac{10^{4030}+10^{2015}-2}{9}=10^{2015}\times \frac{10^{2015}-1}{9}+2\times \frac{10^{2015}-1}{9}=10^{2015}R_{2015}+2\times R_{2015}$$
$2)$ follows from $\frac{1}{2^2}+...+\frac{1}{2015^2}<\sum_{j=2}^{\infty} \frac{1}{j^2}=\frac{\pi^2}{6}-1<\frac{2014}{2015}$
A: 1.
$$9p=\underbrace{99\cdots9}_\text{4030}+\underbrace{99\cdots9}_\text{2015}$$
$$=10^{4030}-1+10^{2015}-1=\left(10^{2015}\right)^2+10^{2015}-2$$
$$=\left(10^{2015}-1\right)\left(10^{2015}+2\right)=\left(10^{2015}-1\right)\left(10^{2015}-1+3\right)$$
$$\therefore p=\frac{10^{2015}-1}{3}\left(\frac{10^{2015}-1}{3}+1\right)$$
$$\therefore n=\frac{10^{2015}-1}{3}=\underbrace{33\cdots3}_\text{2015}$$
2.
$$\sum_{k=2}^{n} \frac1{k^2}<\sum_{k=2}^{n} \frac1{(k-1)k}=\sum_{k=2}^{n} \frac1{k-1}-\frac1{k}=1-\frac{1}{n}=\frac{n-1}{n}$$
A: For 1 you can use the quadratic equation:
$n^2+n-p=0 \rightarrow n=\frac{-1 \pm \sqrt{4 \cdots 4 8 \cdots 89}}{2} $.
Now show $\sqrt{4 \cdots 4 8 \cdots 89}=\underbrace{66\cdots67}_\text{2015}$ and take the positive root to get $n=\underbrace{33\cdots3}_\text{2015}$
