Svaki član niza ( a n ) n ∈ N (a_n)_{n\in\mathbb{N}} ( a n ) n ∈ N
Ako je
a 1 = 1 505 i a 505 = 505 , a_1=\frac1{505}\qquad\text{i}\qquad a_{505}=505, a 1 = 505 1 i a 505 = 505 , odredi a 1010 a_{1010} a 1010
Rješenje Prema uvjetu zadatka za svaki n ≥ 2 n\geq2 n ≥ 2
a n = a n − 1 a n + 1 + a n − 1 + a n + 1 2 2 = ( a n − 1 + a n + 1 ) 2 4 . a_n=\frac{\sqrt{a_{n-1}a_{n+1}}+\frac{a_{n-1}+a_{n+1}}2}{2} =\frac{\left(\sqrt{a_{n-1}}+\sqrt{a_{n+1}}\right)^2}{4}. a n = 2 a n − 1 a n + 1 + 2 a n − 1 + a n + 1 = 4 ( a n − 1 + a n + 1 ) 2 . Dakle,
a n = a n − 1 + a n + 1 2 , \sqrt{a_n}=\frac{\sqrt{a_{n-1}}+\sqrt{a_{n+1}}}{2}, a n = 2 a n − 1 + a n + 1 , pa je niz ( a n ) n ∈ N \left(\sqrt{a_n}\right)_{n\in\mathbb{N}} ( a n ) n ∈ N
Označimo razliku tog aritmetičkog niza s d d d
( 505 − 1 ) d = a 505 − a 1 = 505 − 1 505 = 504 505 , (505-1)d=\sqrt{a_{505}}-\sqrt{a_1} =\sqrt{505}-\frac1{\sqrt{505}} =\frac{504}{\sqrt{505}}, ( 505 − 1 ) d = a 505 − a 1 = 505 − 505 1 = 505 504 , odakle je
d = 1 505 . d=\frac1{\sqrt{505}}. d = 505 1 . Prema tome,
a 1010 = a 1 + ( 1010 − 1 ) d = 1 505 + 1009 505 = 1010 505 = 2 505 . \sqrt{a_{1010}}=\sqrt{a_1}+(1010-1)d =\frac1{\sqrt{505}}+\frac{1009}{\sqrt{505}} =\frac{1010}{\sqrt{505}}=2\sqrt{505}. a 1010 = a 1 + ( 1010 − 1 ) d = 505 1 + 505 1009 = 505 1010 = 2 505 . Konačno,
a 1010 = 4 ⋅ 505 = 2020. a_{1010}=4\cdot505=2020. a 1010 = 4 ⋅ 505 = 2020. 