1. a) (3p) Factorize the number $4 n^{4}+1, n \in \mathbb{N}^{*}$.

b) (4p) Determine the integer part of the number $A=\frac{4}{5}+\frac{8}{65}+\frac{12}{325}+\ldots+\frac{4 n}{4 n^{4}+1}, n \in \mathbb{N}^{*}$.

Gazeta Matematică Nr. 10/2014
Solution: a) $4 n^{4}+1=4 n^{4}+4 n^{2}+1-4 n^{2}=\left(2 n^{2}+1\right)^{2}-4 n^{2}=\left(2 n^{2}+1-2 n\right)\left(2 n^{2}+1+2 n\right)=\left(2 n^{2}-2 n+1\right)\left(2 n^{2}+2 n+1\right)$ b) Applying point a) we generally have: $\frac{4 k}{4 k^{4}+1}=\frac{4 k}{\left(2 k^{2}-2 k+1\right)\left(2 k^{2}+2 k+1\right)}=\frac{\left(2 k^{2}+2 k+1\right)-\left(2 k^{2}-2 k+1\right)}{\left(2 k^{2}-2 k+1\right)\left(2 k^{2}+2 k+1\right)}=$ $=\frac{2 k^{2}+2 k+1}{\left(2 k^{2}-2 k+1\right)\left(2 k^{2}+2 k+1\right)}-\frac{2 k^{2}-2 k+1}{\left(2 k^{2}-2 k+1\right)\left(2 k^{2}+2 k+1\right)}=\frac{1}{2 k^{2}-2 k+1}-\frac{1}{2 k^{2}+2 k+1}, \forall k \in \mathbb{N}^{*}$.

Then $A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+\frac{1}{13}-\frac{1}{25}+\ldots+\frac{1}{2 n^{2}-2 n+1}-\frac{1}{2 n^{2}+2 n+1}=1-\frac{1}{2 n^{2}+2 n+1}$.

Since $0<\frac{1}{2 n^{2}+2 n+1}<1 \Rightarrow 0<A<1 \Rightarrow$ the integer part of $A$ is 0.

## Scoring Guide

| a) $4 n^{4}+1=\left(2 n^{2}-2 n+1\right)\left(2 n^{2}+2 n+1\right)$ | $\mathbf{3 p}$ |
| :--- | :--- |
| b) Applies point a) and determines $A=1-\frac{1}{2 n^{2}+2 n+1}$ | $\mathbf{3 p}$ |
| $0<\frac{1}{2 n^{2}+2 n+1}<1 \Rightarrow 0<A<1 \Rightarrow$ the integer part of $A$ is 0 | $\mathbf{1 p}$ |