First the "verifying" part if fairly straight forward:
$\frac{1}{n^2- n+ 1}+ \frac{1}{n^2+ n+ 1}= \frac{n^2+ n+ 1- (n^2- n+ 1}{(n^2+n+1)(n^2- n+ 1)}= \frac{n^2+ n+ 1- n^2+ n- 1}{(n^2+n)(n^2-n)+ (n^2+ n)+ (n^2- n)+ 1}= \frac{2n}{n^4- n^2+ 2n+ 1}= \frac{2n}{n^4+ n^2+ 1}$
as asserted. The first sum is half the sum of those fractions.
Second, let $\displaystyle f(n)= \frac{1}{n^2- n+ 1}- \frac{1}{n^2+ n+ 1}$ and note that [math]f(1)= 1- \frac{1}{3}, $\displaystyle f(2)= \frac{1}{3}- \frac{1}{7}$, $\displaystyle f(3)= \frac{1}{7}- \frac{1}{13}$, $\displaystyle f(4)= \frac{1}{13}- \frac{1}{21}$, etc. The second term in each f(n) is canceled by the first term in f(n+1). This is a "telescoping series". The only terms that are not canceled in the sum are the first, 1, and the last, $\displaystyle =\frac{1}{N^2+ N+ 1}$. And $\displaystyle \frac{1}{2}\left(1- \frac{1}{N^2+ N+ 1}\right)= \frac{N^2+ N+ 1- 1}{N^2+ N+ 1}= \frac{N^2+ N}{N^2+ N+ 1}$.
First the "verifying" part if fairly straight forward:
$\frac{1}{n^2- n+ 1}+ \frac{1}{n^2+ n+ 1}= \frac{n^2+ n+ 1- (n^2- n+ 1}{(n^2+n+1)(n^2- n+ 1)}= \frac{n^2+ n+ 1- n^2+ n- 1}{(n^2+n)(n^2-n)+ (n^2+ n)+ (n^2- n)+ 1}= \frac{2n}{n^4- n^2+ 2n+ 1}= \frac{2n}{n^4+ n^2+ 1}$
as asserted. The first sum is half the sum of those fractions.
Second, let $\displaystyle f(n)= \frac{1}{n^2- n+ 1}- \frac{1}{n^2+ n+ 1}$ and note that $\displaystyle f(1)= 1- \frac{1}{3}$, $\displaystyle f(2)= \frac{1}{3}- \frac{1}{7}$, $\displaystyle f(3)= \frac{1}{7}- \frac{1}{13}$, $\displaystyle f(4)= \frac{1}{13}- \frac{1}{21}$, etc. The second term in each f(n) is canceled by the first term in f(n+1). This is a "telescoping series". The only terms that are not canceled in the sum are the first, 1, and the last, $\displaystyle =\frac{1}{N^2+ N+ 1}$. And $\displaystyle \frac{1}{2}\left(1- \frac{1}{N^2+ N+ 1}\right)= \frac{N^2+ N+ 1- 1}{2(N^2+ N+ 1)}= \frac{N^2+ N}{2(N^2+ N+ 1)}$.