ПМИ arcctg

[Гост]

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[Добромир Глухаров]

$ctg(\alpha+\beta)=\frac{ctg\alpha ctg\beta-1}{ctg\alpha+ctg\beta}$

$\alpha+\beta=arcctga+arcctgb=arcctg\frac{ab-1}{a+b}$

$arcctg2+arcctg8+\cdots+arcctg2n^2+arcctg2(n+1)^2=arcctg\frac{n+1}{n}+arcctg2(n+1)^2=arcctg\frac{\frac{2(n+1)^3}{n}-1}{\frac{n+1}{n}+2(n+1)^2}=$

$=arcctg\frac{2(n+1)^3-n}{(n+1)(1+2n(n+1))}=arcctg\frac{(n+2)(2n^2+2n+1)}{(n+1)(2n^2+2n+1)}=arcctg\frac{n+2}{n+1}$ - тъждество и за $n+1$.