数列 $\{a_n\}$ が与えられた漸化式で定義されるとき、第3項から第5項 $(a_3, a_4, a_5)$ をそれぞれ求めます。 (1) $a_1 = 1, a_{n+1} = 4a_n + 1$ (2) $a_1 = -1, a_{n+1} = a_n + 2n$代数学数列漸化式2025/6/251. 問題の内容数列 {an}\{a_n\}{an} が与えられた漸化式で定義されるとき、第3項から第5項 (a3,a4,a5)(a_3, a_4, a_5)(a3,a4,a5) をそれぞれ求めます。(1) a1=1,an+1=4an+1a_1 = 1, a_{n+1} = 4a_n + 1a1=1,an+1=4an+1(2) a1=−1,an+1=an+2na_1 = -1, a_{n+1} = a_n + 2na1=−1,an+1=an+2n2. 解き方の手順(1) a1=1,an+1=4an+1a_1 = 1, a_{n+1} = 4a_n + 1a1=1,an+1=4an+1a2=4a1+1=4(1)+1=5a_2 = 4a_1 + 1 = 4(1) + 1 = 5a2=4a1+1=4(1)+1=5a3=4a2+1=4(5)+1=21a_3 = 4a_2 + 1 = 4(5) + 1 = 21a3=4a2+1=4(5)+1=21a4=4a3+1=4(21)+1=85a_4 = 4a_3 + 1 = 4(21) + 1 = 85a4=4a3+1=4(21)+1=85a5=4a4+1=4(85)+1=341a_5 = 4a_4 + 1 = 4(85) + 1 = 341a5=4a4+1=4(85)+1=341(2) a1=−1,an+1=an+2na_1 = -1, a_{n+1} = a_n + 2na1=−1,an+1=an+2na2=a1+2(1)=−1+2=1a_2 = a_1 + 2(1) = -1 + 2 = 1a2=a1+2(1)=−1+2=1a3=a2+2(2)=1+4=5a_3 = a_2 + 2(2) = 1 + 4 = 5a3=a2+2(2)=1+4=5a4=a3+2(3)=5+6=11a_4 = a_3 + 2(3) = 5 + 6 = 11a4=a3+2(3)=5+6=11a5=a4+2(4)=11+8=19a_5 = a_4 + 2(4) = 11 + 8 = 19a5=a4+2(4)=11+8=193. 最終的な答え(1) a3=21,a4=85,a5=341a_3 = 21, a_4 = 85, a_5 = 341a3=21,a4=85,a5=341(2) a3=5,a4=11,a5=19a_3 = 5, a_4 = 11, a_5 = 19a3=5,a4=11,a5=19