Қалалық Жәутіков олимпиадасы 8 сынып, 2015 жыл

Теңдікті дәлелдеңдер: $\sqrt {1 + \dfrac{1}{{{1^2}}} + \dfrac{1}{{{2^2}}}} + \sqrt {1 + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}}} + \ldots + \sqrt {1 + \dfrac{1}{{{{2014}^2}}} + \dfrac{1}{{{{2015}^2}}}} = 2015 - \dfrac{1}{{2015}}.$

Комментарий/решение

$\sqrt{1+\dfrac{1}{x^2} + \dfrac{1}{(x+1)^2}} = \dfrac{1+x + x^2}{x+x^2}$

Получим сумму

$S=\dfrac{3}{2}+\dfrac{7}{6}+\dfrac{13}{12}+\dfrac{21}{20}+\dfrac{31}{30}+...+\dfrac{4058211}{4058210}$

$S=2014+\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{4058210}$

$S = 2014+(1-\dfrac{1}{2})+(\dfrac{1}{2}-\dfrac{1}{3})+(\dfrac{1}{3}-\dfrac{1}{4})+( \dfrac{1}{4}-\dfrac{1}{5})+...+\dfrac{1}{2014}-\dfrac{1}{2015}=2015-\dfrac{1}{2015}$