첫째항이 1이고 공차가 양수인 등차수열 { a n } \left\{a_n\right\} { a n } n n n y = a n y=a_n y = a n y = log 2 x y=\log_2x y = log 2 x P n \mathrm{P}_n P n P n \mathrm{P}_n P n x x x x x x y = x 2 y=x^2 y = x 2 A n , B n \mathrm{A}_n,\,\mathrm{B}_n A n , B n A n B n P n + 1 \mathrm{A}_n\mathrm{B}_n\mathrm{P}_{n+1} A n B n P n + 1 T n T_n T n
다음은 모든 자연수 n n n O A n ‾ : O A n + 1 ‾ = 1 : 4 \overline{\mathrm{OA}_n}:\overline{\mathrm{OA}_{n+1}}=1:4 OA n : OA n + 1 = 1 : 4 ∑ n = 1 5 T n \displaystyle\sum_{n=1}^{5}T_n n = 1 ∑ 5 T n O \mathrm{O} O
a n = log 2 x a_n=\log_2x a n = log 2 x x = 2 a n x=2^{a_n} x = 2 a n A n \mathrm{A}_n A n x x x 2 a n 2^{a_n} 2 a n O A n ‾ = 2 a n , O A n + 1 ‾ = 2 a n + 1 \overline{\mathrm{OA}_n}=2^{a_n},\,\quad \overline{\mathrm{OA}_{n+1}}=2^{a_{n+1}} OA n = 2 a n , OA n + 1 = 2 a n + 1 O A n ‾ : O A n + 1 ‾ = 1 : 4 \overline{\mathrm{OA}_n}:\overline{\mathrm{OA}_{n+1}}=1:4 OA n : OA n + 1 = 1 : 4
2 a n : 2 a n + 1 = 1 : 4 2^{a_n}:2^{a_{n+1}}=1:4 2 a n : 2 a n + 1 = 1 : 4 이다. 그러므로 등차수열 { a n } \left\{a_n\right\} { a n } (가) \boxed{\text{(가)}} ( 가 ) B n \mathrm{B}_n B n ( 2 a n , 4 a n ) \left(2^{a_n},\,4^{a_n}\right) ( 2 a n , 4 a n )
T n = 1 2 × A n B n ‾ × A n A n + 1 ‾ = 1 2 × 4 a n × ( 2 a n + 1 − 2 a n ) = (나) × 8 a n \begin{aligned}T_n&=\frac12\times\overline{\mathrm{A}_n\mathrm{B}_n}\times\overline{\mathrm{A}_n\mathrm{A}_{n+1}}\\[6pt]&=\frac12\times4^{a_n}\times\left(2^{a_{n+1}}-2^{a_n}\right)\\[6pt]&=\boxed{\text{(나)}}\times8^{a_n}\end{aligned} T n = 2 1 × A n B n × A n A n + 1 = 2 1 × 4 a n × ( 2 a n + 1 − 2 a n ) = ( 나 ) × 8 a n 이다. 수열 { a n } \left\{a_n\right\} { a n }
∑ n = 1 5 T n = (다) × ( 2 30 − 1 ) \sum_{n=1}^{5}T_n=\boxed{\text{(다)}}\times\left(2^{30}-1\right) n = 1 ∑ 5 T n = ( 다 ) × ( 2 30 − 1 ) 이다.
위의 (가), (나), (다)에 알맞은 수를 각각 p , q , r p,\,q,\,r p , q , r p q × r \dfrac{p}{q\times r} q × r p
