Найди сумму 1/1+√2+1/√2+√3+...1/√2004+√2005

1/(1+√2) = (√2–1)/((√2–1)(√2+1)) = (√2–1)/(2–1) = √2–1, 1/(√2+√3) = (√3–√2)/((√3–√2)(√3+√2)) = (√3;–√2)/(3–2) = √3–√2, . . 1/(√2004+√2005) = (√2005–√2004)/((√2005–√2004)(√2005+√2004)) = (√2005–√2004)/(2005–2004) = √2005–√2004. Сумма равна √2–1+√3–√2+…+√2005–√2004 = √2005–1.